**Heat Capacity**

**Heat Capacity, C, of a body is defined as the amount of heat (Q) required to raise its temperature (θ) by one degree, without going through a change of state.**- Amount of heat needed to raise the temperature of an object depends on the mass of the object.
- Heat capacity also depends on the material of the object. Some materials are harder to heat up than others. The molecules in a liquid such as water require more energy to move faster than copper atoms in a solid. So, in order to record 1°C increase in temperature, liquids would require more heat energy than solids.
- SI. unit of heat capacity is joule per kelvin (J K
^{-1}) or joule per degree Celsius (J °C^{-1}).

$C = \frac{Q}{\Delta \theta}$ , where

C = heat capacity (JK^{-1}, J°C^{-1})

Q = heat or thermal energy absorbed or released (J)

Δθ = change in temperature (K or °C)

Heat Capacity | Time to cool down/heat up | Reason |
---|---|---|

High | Longer | Need to lose more energy (cooling) or absorb more energy (heating) |

Low | Shorter | Need to lose less energy (cooling) or absorb less energy (heating) |

**Worked Example:**

In a simple experiment, 100 g of water requires 12 600 J of heat to raise it from 30 °C to 60 °C.

i) Find the heat capacity of 100 g of water.

ii) Find the heat capacity of 1000 g of water.

iii) Find the heat needed to raise 1000 g of water from 30 °C to 40 °C.

**Show/Hide Answer**

i) $C = \frac{Q}{\Delta \theta} = \frac{12600}{30}$

C = 420 JK^{-1}

ii) Since 1000 g of water has 10 times the mass of 100 g of water, the heat capacity of 1000 g of water = 10 times the heat capacity of 100 g of water.

Hence, C for 1000 g of water = 4200 JK^{-1}

iii) Heat needed, $Q = C \Delta \theta = 4200 (40 – 30) = 42000 J$

In the question, The answer is given in J/K-1 but the values in statement are in Degree Celsius.

The fundamental intervals [the difference between the lower and upper fixed points] between the Kelvin and Celsius scales are the same, that’s the two temperature scales are are devided into 100 equal parts. This means that temperature difference between both the two Scales will always be the same and in solving problems concerning Heat Capacity we are only interested in the temperature change not the Scales.