Quantity of Heat

Quantity of Heat

 –Heat is a form of energy that flows from one body to another due to temperature differences between them.

Heat capacity

Heat capacity is defined as the quantity of heat required to raise the temperature of a given mass of a substance by one degree Celsius or one Kelvin. It is denoted by C.
Heat capacity, C = heat absorbed, Q / temperature change θ.
The units of heat capacity are J / °C or J / K.

Specific heat capacity

S.H.C of a substance is the quantity of heat required to raise the temperature of 1 kg of a substance by 1 °C or 1 K. It is denoted by ‘c’, hence,
c = Q / m θ where Q – quantity of heat, m – mass and θ – change in temperature. The units for ‘c’ are J kg^-1 K^-1. Also Q = m c θ.
Example 1
1. A block of metal of mass 1.5 kg which is suitably insulated is heated from 30 °C to 50 °C in 8 minutes and 20 seconds by an electric heater coil rated 54 watts. Find;
a) The quantity of heat supplied by the heater
b) The heat capacity of the block
c) Its specific heat capacity
Solution

a) Quantity of heat = power × time = P t
= 54 × 500 = 27,000 J
b) Heat capacity, C = Q / θ = 27,000 / (50 – 30) = 1,350 J K^-1
c) Specific heat capacity, c = C / m = 1,350 / 1.5 = 900 J Kg^-1 K^-1

Example 2
If 300 g of paraffin is heated with an immersion heater rated 40 W, what is the temperature after 3 minutes if the initial temperature was 20 °C? (S.H.C for paraffin = 2,200 J Kg^-1 K^-1).
Solution

Energy = P t = m c θ = Q = quantity of heat.
P t = 40 ×180 = 7,200 J
m = 0.30 kg c = 2,200, θ = ..?
Q = m c θ, θ = Q / m c = 7,200 / (0.3 × 2,200) = 10.9 °C
Final temperature = 20°C + 10.9°C = 30.9°C

Example 3
A piece of copper of mass 60 g and specific heat capacity 390 J Kg^-1 K^-1 cools from 90 °C to 40 °C. Find the quantity of heat given out.
Solution

Q = m c θ, = 60 × 10^-3 × 390 × 50 = 1,170 J.

Determination of specific heat capacity

– A calorimeter is used to determine the specific heat capacity of a substance.
– This uses the principle of heat gained by a substance is equal to the heat lost by another substance in contact with each other until equilibrium is achieved.
– Heat losses in calorimeter are controlled such that no losses occur or they are very minimal.

Example 4
1. A 50 W heating coil is immersed in a liquid contained in an insulated flask of negligible heat capacity. If the mass of the liquid is 10 g and its temperature increases by 10 °C in 2 minutes, find the specific heat capacity of the liquid.
Solution

Heat delivered (P t) = 50 × 2 × 60 = 2,400 J
Heat gained = 0.1 × c × 10 J
Therefore ‘c’ = 2,400 / 0.1 × 10 = 2,400 J Kg^-1 K^-1

Example 5
A metal cylinder mass 0.5 kg is heated electrically. If the voltmeter reads 15V, the ammeter 3A and the temperatures of the block rises from 20 °C to 85 °C in ten minutes. Calculate the specific heat capacity of the metal cylinder.
Solution

Heat gained = heat lost, V I t = m c θ
15 × 3 × 10 × 60 = 0.5 × c × 65
c = (15 × 3 × 600)/ 0.5 × 65 = 831 J Kg^-1 K^-1

Fusion and latent heat of fusion

– Fusion is the change of state from solid to liquid. Change of state from liquid to solid is called solidification. Latent heat of fusion is the heat energy absorbed or given out during fusion.
– Specific latent heat of fusion of a substance is the quantity of heat energy required to change completely 1 kg of a substance at its melting point into liquid without change in temperature.
– It is represented by the symbol (L), we use the following formula,
Q = m Lf
– Different substances have different latent heat of fusion.

Factors affecting the melting point

1. Pressure
2. Dissolved substances

– Specific latent heat of vaporization is the quantity of heat required to change completely 1 kg of a liquid at its normal boiling point to vapour without changing its temperature. Hence
Q = m L v
The SI unit for specific latent heat of vaporization is J / Kg.

Example 6
An immersion heater rated 600 W is placed in water. After the water starts to boil, the heater is left on for 6 minutes. It is found that the mass of the water had reduced by 0.10 kg in that time. Estimate the specific heat of vaporization of steam.
Solution

Heat given out by the heater = P t = 600 × 6 × 60
Heat absorbed by steam = 0.10 × L v
Heat gained = heat lost, therefore, 600 × 6 × 60 = 0.10 × L v = 2.16 × 10⁶J / Kg

Evaporation

Factors affecting the rate of evaporation

• Temperature
• Surface area
• Draught (hot and dry surrounding)
• Humidity

Comparison between boiling and evaporation

Evaporation
1. Takes place at all temperature.
2. Takes place on the surface (no bubbles formed)
3. Decrease in atmospheric pressure increases the rate.

Boiling
1.Takes place at a specific temperature
2.Takes place throughout the liquid ( bubbles formed)
3.Decreases as atmospheric pressure lowers

Applications of cooling by evaporation

a) Sweating
b) Cooling of water in a porous pot
c) The refrigerator