** Heat:** A form of energy that gives the sensation of warmth is called heat OR The sum of kinetic energy of the molecules present in a body called heat.

** Unit of Heat:** Its unit are Joule and calorie and S.I. unit is Joule (J).

__Relation of Joule and calorie__

4.2 Joule = 1 Calorie

** Device to measure heat**: Heat is measured with calorimeter.

Factors on which heat depends:

- Mass of substance
- Temperature of substance

**Temperature:** The degree of hotness or coldness of body is called temperature OR The average kinetic energy of the molecules present in body is called temperance.

**a) What is the total kinetic energy of given substance?**

1J+3J+5J+7J = 16J

**b) What is the heat energy in given substance?**

16J heat energy.

**c) What is the average kinetic energy of substance?**

Average kinetic energy = (1+3+5+7)/4 = 16/4 = 4J

**d) What is the temperature?**

4J

** Unit of temperature:**The unit of temperature are:

- Degree celsius (°C)
- Degree Fahrenheit (°F)
- Kelvin (K)

**Device to measure temperature. **Thermometer is used to measure temperature.

**Q) Write 4 differences between heat and temperature**

Heat | Temperature |

It is the sum of kinetic energy of molecules present in body. | It is the average of kinetic energy of molecules present in body. |

Its S.I. unit is Joule (J). | Its SI unit is Kelvin (K). |

It is measured with caloriemeter. | It is measured with Thermometer. |

It is cause. | It is effect |

It flows. | It doesn’t flows. |

__Relation between different units of temperature__

`(C-0)/100 = (F-32)/180 = (F-273)/100`

**Q. Convert 37°C into Fahrenheit scale.**

C = 37°C

F = ?

We have,

(C-0)/100 = (F-32)/180

37 x I80 = 10O x F – 3200

100 x F = 6600+3200

F= 98.6°F

Hence, 37°C = 98.6°F

** Thermometer:** Thermometer the device which is used to measure temperature of body.

** Principle of thermometer: **The principle of thermometer states “the substances expand on heating and contract on cooling.”

** Thermometric liquid:** The liquid which is used in thermometer is called thermometric liquid. Most commonly used thermometric liquids are mercury and alcohol.

__Mercury as thermometric liquid__

- It is shiny in appearance.
- It is non-sticky to the walls of container.
- It can measure very wide range of temperature as its melting point is -39°C and boiling point point is 357°C.
- It has uniform rate of expansion and contraction.
- It is very sensitive to heat because its specific heat capacity is very low.

__Alcohol as thermometric liquid__

- It is cheap and easily available.
- It is transparent so that it can be coloured.
- It can measure wide range of temperature ie from -117°C to 78°C.
- The rate of expansion and contraction is 6 times more than that of mercury.

**Q. Write melting point and boiling point of alcohol and mercury **

Melting Point | Boiling Point | |

Mercury | -39°C | 357°C |

Alcohol | -117°C | 78°C |

**Q. Which thermometers is suitable to measure temperature at very hot places. Why?**

Mercury thermometer is suitable to measure temperature at very hot places because it can measure high temperature ie up to 357°C.

**Q. Which thermometer is suitable to measure the temperature at very cold place. why?**

Alcohol thermometer is suitable to measure temperature at very cold place because it can measure very low temperature i.e. up to -117°C.

__Types of thermometer__

- Clinical thermometer/Doctor’s thermometer
- Laboratory thermometer/Simple thermometer
- Digital thermometer
- Maximum-minimum thermometer

__A. Clinical thermometer__

The thermometer which is used to measure the human body temperature is called clinical thermometer.

Features of clinical thermometer

- It is prismatic in shape.
- It is calibrated from 35°C – 42°C or 94°F – 108°F
- Constriction (kink) is kept near its bulb.

**Q. Why is kink kept in clinical thermometer?**

Kink is kept in clinical thermometer to prevent back flow of mercury in capillary tubes so that correct temperature of the body can be measured even when the thermometer is taken away from the body.

__B. Laboratory thermometer__

The thermometer which is used to measure the temperature of different substances during experiment is called Laboratory thermometer. It is also called simple thermometer.

**Q. Write the differences between clinical and laboratory thermometer.**

Clinical thermometer | Laboratory thermometer |

It is used to measure human body temperature. | It is used to measure temperature of different substrate during experiment. |

Kink is kept in it. | Kink is absent. |

It is calibrated from 35°C to 42°C. | It is calibrated from -10° to 110°C. |

It is called doctors thermometer | It called simple thermometer. |

__C. Digital thermometer__

The advance thermometer in which thermistor is used as thermometric substance and the temperature is displayed in digit with alarm beep is called digital thermometer.

__D. Maximum-minimum Thermometer.__

The thermometer that measure maximum and minimum temperature of a particular place of 24 hours is called maximum minimum thermometer.

__Working of maximum-minimum thermometer__

When temperature of atmospheric pressure increases, alcohol in minimum side expand which pushes the index of the side downwards so that mercury level side increases. The increased level of mercury pushes index of maximum side upwards and gives maximum temperature.

When the temperature of atmosphere decreases, the alcohol in minimum side contracts so that index of this side rises up as mercury of maximum side falls downward. This shows minimum temperature of a place.

In this way, a maximum and minimum thermometer measures maximum and minimum temperature of place.

** Calorie heat: **The amount of heat energy required to change the temperature of 1gm of water by 1°C is called 1 calorie heat.

** Specific heat capacity (s):** The amount of heat energy required to change temperature of 1 kg mass by 1°C is called specific heat capacity..

Unit of specific heat capacity is J/kg°C or Jkg^{-1}°C^{-1}

**a. Specific heat capacity of water is 4200 J/kg°C. What does it mean?**

It means that 4200J heat energy is required to change temperature of 1 kg water by 1°C.

**b. Specific heat capacity of iron is 470 J/kg°C. What does it mean?**

It means that 470J heat energy is required to change temperature of 1kg iron by 1°C.

**c. Why is water used in the engines of automobiles?**

The specific heat capacity of water is maximum so that it can absorb large amount of heat from the engine of automobiles and keeps them cool without raising its own temperature much. Hence, water is used in engines of automobiles.

**d. Why is water used in hot water bag?**

The specific heat capacity of water is maximum so that it remains hotter for longer duration and hot water bag can be used for long period of time. Hence, water is used in hot water bag.

**e. Days are very hotter and night are much colder in desert. Why?**

Desert mainly contain sand which has very low specific heat capacity so that sand gets heated very quickly to high temperature during the day and get cooled down very quickly to low temperature during night. Hence, days are very hot and nights are much cooler in desert.

**f. Well or spring water is warmer in morning and colder in day. why?**

Specific heat capacity of water is maximum so that its temperature remains almost same. The temperature of atmosphere is lower than the temperature of water in well spring in morning so that it is felt warmer but temperature of atmosphere is higher than temperature of water in well spring during day so that it is felt colder. Hence, well spring water are warmer in morning and calder in day.

__Heat equation__

We have heat energy gained or lost by a body is directly proportional to mass and its change in temperature. If heat energy gained or lost by a body of mass ‘m’ is ‘Q’ then its change in temperature (t_{2}-t_{1}) is dt

Now,

Q ∝ m …………. I

Q ∝ dt …………. II

Combining equation I and II we get,

Q ∝ m.dt

Q = s.m.dt

where s is the specific heat capacity of a body which is constant for that body.

__Determining unit of specific heat capacity.__

We have,

Q = s.m.dt

s = Q / m.dt

Now, S = J/kg°C or Jkg^{-1}°C^{–}

Hence, unit of S is J/kg°C or Jkg^{-1}°C^{–}

**Q) Three substance with different specific heat capacities are given below in the table. Answer the questions on the basis of it.**

A | 900 J/kg°C |

B | 1600 J/kg°C |

C | 800 J/kg°C |

**a. The specific heat capacity of B is 1600 J/kg°C What does it mean?**

It means that 1600 J is required to change temperature of a substance B of 1 kg mass by 1°C.

**b. If equal amount of heat energy is given to equal mass of them, which one will have highest temperature. why?**

If equal amount of heat energy is given to equal mass of them, C will have the highest temperature because it has the lowest specific heat capacity among them so its change in temperature occurs maximum.

**c. Which one of them will require more heat if their temperature is raised equally. Why?**

B will require more heat energy because its specific heat capacity is highest.

**d. If equal mass of same temperature is kept on wax slab, which will penetrate it deeper? write.**

If equal mass of same temperature are kept on wax, B will penetrate deeper because it has highest be specific heat capacity so it remains hot for a long time.

** Principle of calorimetry:** It states that, “When two bodies are in thermal equilibrium heat lost by a body is equal to heat gained by another body.”

i.e. Heat lost = Heat gain

`s2m2(t1-t) = s2m2(t-t2)`

Where, ’t’ is common temperature.

__Solved numerical__

**a. 19000J heat energy is given to water of mass 5kg at 10°C. What will be its final temperature. (specific heat Capacity of water is 4200J/kg°C)**

Q = 19000J

s = 4200J/kg°C

m = 5kg

dt = ?

We know,

Q = s.m.dt

19000 = 4200 x 5 x dt

dt = 0.9°C

Again,

t_{1} = 10°C

t_{2} = ?

We have,

dt = t_{2} – t_{1}

0.9 = t_{2 }– 10

t_{2 }= 10.9°C

Hence, Its final temperature will be to 9°C.

**b. How much heat energy should be supplied to 2kg iron to heat it from 10°C to 180°C. The specific heat capacity of iron is 460 J/kg°C.**

m = 2kg

t_{2} = 180°C

t_{1} = 10°C

dt = t_{2} – t_{1} = 180-10 = 170°C

s = 460J/kg°C

Q = ?

We know

Q = s.m.dt

Q = 460 x 2 x 170

Q = 156400 J

Hence, 156400J should be supplied to the iron.

**c. 20 KJ heat is given to a body of mass 10kg at temperature 10°C to heat it upto 90°C. Calculate the specific heat capacity of the body.**

m = 10kg

t_{2} = 90°C

t_{1} = 10°C

dt = t_{2} – t_{1} = 90-10 = 80°C

Q = 20KJ = 20 x 1000 = 20000J

s = ?

We know

Q = s.m.dt

20000 = 10 x s x 80

s = 25 J/kg°C

Hence, its specific heat capacity is 25 J/kg°C.

**d. Krishma mixes 5 kg of water at 60°C in 10 kg of water at 10°C. Find final temperature of mixture water.**

Let the final temperature be t°C

For hot water,

m_{1} = 5kg

t_{1} = 60°C

s_{1} = 4200J/kc°C

dt_{1} = t_{1} – t = (60-t)°C

Heat lost = s_{1}m_{1}dt_{1 }= 4200 x 5 x (60-t)

For cold water,

m_{2} = 10kg

t_{2} = 10°C

s_{2} = 4200J/kg°C

dt_{2} = t – t_{2} = (t-10)°C

Heat gained = s_{2}m_{2}dt_{2 }= 4200 x 10 x (t-10)

Now, we have

Heat lost = Heat gained

4200 x 5 x (60-t) = 4200 x 10 x (t-10)

60-t = 2 t-20

t = 80/3 = 26.67°C

Hence, mixture temperature is 26.67°C

**e. Nancy plunged an iron of mass 2kg at 200°C into water of mass 20kg at 20°C. Calculate final temperature of both.**

Let the final temperature be t°C

For hot iron,

m_{1} = 2kg

t_{1} = 200°C

s_{1} = 460J/kc°C

dt_{1} = t_{1} – t = (200-t)°C

Heat lost = s_{1}m_{1}dt_{1 }= 460 x 2 x (200-t)

For cold water,

m_{2} = 20kg

t_{2} = 20°C

s_{2} = 4200J/kg°C

dt_{2} = t – t_{2} = (t-20)°C

Heat gained = s_{2}m_{2}dt_{2 }= 4200 x 20 x (t-20)

Now, we have

Heat lost = Heat gained

920 x (200-t) = 84000 x (t-20)

200-t = 91.3(t-20)

t = 2026 / 92.3 = 21.95°C

Hence, final temperature of both is 21.95°C

**f. How much heat is required to raise temperature of 10kg of water by 10°C?**

s = 4200J/kg°C

m = 10kg

dt = 10°C

Q = ?

We know,

Q = s.m.dt = 4200 x 10 x 10 = 420000 J

Hence, 420000 J is required.

**g. Calculate the amount of heat required to heat an iron ball of mass 450g from 35°C to 52°C. If its specific heat capacity is 470 J/kg°C.**

m = 450g = 450/1000 = 0.45 kg

t_{1} = 35°C

t_{2} = 52°C

s = 470J/kg°C

dt = 52-35 = 17°C

Q = ?

We know,

Q = s.m.dt = 470 x 0.45 x 17 = 3595.5 J

**h. Calculate amount of heat lost by our body when we have 250g cold drinks at 4°C if its specific heat capacity is 3800J/kg°C.**

m = 250g = 250/1000 = 0.25 kg

t_{1} = 37°C

t_{2} = 4°C

s = 3800J/kg°C

dt = 4-75 = -33°C

Q = ?

We know,

Q = s.m.dt = 3800 x 0.25 x (-33) = -31350J

Hence, 31350J heat is lost by out body.

**i. Calculate final temperature of water if 10 kg of water at 80°C is mixed with 21 kg of water at 15°C.**

Let the final temperature be t°C

For hot water,

m_{1} = 10kg

t_{1} = 80°C

s_{1} = 4200J/kc°C

dt_{1} = t_{1} – t = (80-t)°C

Heat lost = s_{1}m_{1}dt_{1 }= 4200 x 10 x (80-t)

For cold water,

m_{2} = 21kg

t_{2} = 15°C

s_{2} = 4200J/kg°C

dt_{2} = t – t_{2} = (t-15)°C

Heat gained = s_{2}m_{2}dt_{2 }= 4200 x 21 x (t-15)

Now, we have

Heat lost = Heat gained

4200 x 10 x (80-t) = 4200 x 21 x (t-15)

800-10t = 21t-315

31t = 1115

t = 35.97 °C

Hence, final temperature of both is 35.97 °C

**j. Hari drinks 250ml of a cold drink at 11°C. Calculate total heat lost from his body if specific heat capacity of cold drink is 3900 J/kg°C.**

m = 250l = 250/1000 = 0.25 l = 0.25 kg [Since 1 L =1 kg]

t_{1} = 37°C

t_{2} = 11°C

s = 3900J/kg°C

dt = 11-37 = -26°C

Q = ?

We know,

Q = s.m.dt = 3900 x 0.25 x (-26) = -25360J

Hence, 25360J heat is lost by his body.

**k. Simran drinks 75ml of tea at 63°C. Calculate total heat gained by her body.**

m = 75ml = 75/1000 = 0.075 l = 0.075 kg [Since 1 L =1 kg]

t_{1} = 37°C

t_{2} = 63°C

s = 3200J/kg°C

dt = 63-37 = 26°C

Q = ?

We know,

Q = s.m.dt = 3200 x 0.075 x 26 = 6240J

Hence, 6240J heat is gained by her body.

**l. A 1 KW heater is used to raise temperature of 10 kg of body from 35°C to 40°C. If it takes 2 minutes for work, calculate specific heat capacity of body.**

Power(P) = 1 KW =1 x 1000 W = 1000W

Time(t) = 2x 60 sec = 120sec

Power = Work done / time

1000 = W / 120

W = 120000 J

Since, Work done = Energy

Energy = 120000J

i.e. Q= 120000

m = 10 kg

t_{1} = 35°C

t_{2} = 40°C

dt = 40-35 = 5°C

Q = 120000J

s = ?

We know,

Q = s.m.dt

120000 = s x 10 x 5

S = 2400 J/kg°C

Hence, specific heat capacity of body is 2400 J/kg°C

**m. When 200 ml of water at 90°c and 100 ml of water at 15° C ore mixed together, what will be final temperature?**

Let the final temperature be t°C

For hot water,

m_{1} = 200ml = 200/1000 = 0.2 l = 0.2 kg

t_{1} = 90°C

s_{1} = 4200J/kc°C

dt_{1} = t_{1} – t = (90-t)°C

Heat lost = s_{1}m_{1}dt_{1 }= 4200 x 0.2 x (90-t)

For cold water,

m_{2} = 100ml = 100/1000 = 0.1 l = 0.1 kg

t_{2} = 15°C

s_{2} = 4200J/kg°C

dt_{2} = t – t_{2} = (t-15)°C

Heat gained = s_{2}m_{2}dt_{2 }= 4200 x 0.1 x (t-15)

Now, we have

Heat lost = Heat gained

4200 x 0.2 x (80-t) = 4200 x 0.1 x (t-15)

18-0.2t = 0.1t-1.5

0.3t = 19.5

t = 19.5/0.3 = 65°C

Hence, final temperature of both is 65°C