Specific Heat Capacity

Specific Heat Capacity (kJ/kg/°C) The specific heat is the amount of heat (measured in kJ) per unit mass (measured in Kg) required to raise the temperature by one degree Celsius. The relationship between heat and temperature change is usually expressed in the form shown below where c is the specific heat. Where: –Q = Heat added –c = Specific heat –m = Mass –∆T = The rise in temperature (t final - t initial ) Q = cm∆T

The relationship does not apply if a phase change is encountered, because the heat added or removed during a phase change does not change the temperature. The specific heat of water is; –1 calorie/gram °C –4.186 joule/gram °C which is higher than any other common substance. As a result, water plays a very important role in temperature regulation. The specific heat per gram for water is much higher than that for a metal.

Given that joule/gram °C answer the following questions. –1. How much energy would it take to raise the temperature of 1 litre of water from 20°C to 100°C. –2. How much energy would it take to raise the temperature of 0.5 litre of water from 24°C to 100°C. –3. How much energy would it take to raise the temperature of 95 litres of water from 40°C to 85°C.

1. How much energy would it take to raise the temperature of 1 litre of water from 20°C to 100°C. Q = cmΔT Q = Specific Heat Capacity x Mass x (t final – t initial) Q = 4182 x 1 x (100°C - 20°C) Q = 4182 x 80 Q = J

2. How much energy would it take to raise the temperature of 0.5 litre of water from 24°C to 100°C. Q = cmΔT Q = Specific Heat Capacity x Mass x (t final – t initial) Q = 4186 x 0.5 x (100°C - 24°C) Q = 2093 x 76 Q = J Q = KJ

3. How much energy would it take to raise the temperature of 95 litres of water from 40°C to 85°C. Q = cmΔT Q = Specific Heat Capacity x Mass x (t final – t initial) Q = 4186 x 95 x (85°C - 40°C) Q = x 45 Q = J Q = MJ

Specific heat of water = 4186 j/kg/ºC 1 kWh is equal to joule. 4.2p kWh for gas. Ambient temperature is 16ºC A solar thermal system has a capacity of 100l; it heats the water to 45ºC the boiler heats the water to 87ºC. Given that Q = cmΔT How much money would you save over a 7 day period if the heating is used 6 hours per day.

First heat differential calculation Q = cmΔT Q = x 100,000 x 71 Q = x 71 Q = J Second heat differential calculation Q = cmΔT Q = x 100,000 x 42 Q = x 42 Q = J Convert joules to Kwh ÷ Kwh Convert joules to Kwh ÷ Kwh Time Kwh x 6 x Kwh Time Kwh x 6 x Kwh Cost Kwh x 4.2p p £14.56p Cost Kwh x 4.2p p £8.62p Saving in Cost £14.56p - £8.62p £5.94p