- Calorimetry means measurement of Heat.

- Calorimeter is the device used to measure heat and it is cylindrical vessel made of copper and provided by a stirrer and a lid.

- This vessel is kept in a wooden block to isolate it thermally from surroundings.A thermometer is used to measure the temperature of the content in the calorimeter.

- When bodies at different temperature are mixed in a calorimeter,they exchange heat with each other.

- Bodies at higher temperature loose heat while bodies at low temperature gain heat.Contents of the calorimeter is continuously stirred to keep temperature of contents uniform

- Thus principle of calorimetry states that the total heat given by hot objects is equal to the total heat received by cold objects.

$\text{Heat Lost by hot objects} =\text {Heat Gained by cold object}$

A sphere of aluminium of 0.047 kg placed for sufficient time in a vessel containing boiling water, so that the sphere is at 100 °C. It is then immediately transferred to 0.14 kg copper calorimeter containing 0.25 kg of water at 20 °C. The temperature of water rises and attains a steady state at 23 °C. Calculate the specific heat capacity of aluminium.?

Specific heat capacity of water (s

Specific heat capacity of copper calorimeter (s

Here we would be using the principle of calorimetry i.e

$ \text {heat given by an aluminium sphere} =\text { heat absorbed by the water} \\ + \text {heat absorbed by copper calorimeter}$.

Given

Mass of aluminium sphere ($m_1$) = 0.047 kg

Initial temp. of aluminium sphere = 100 °C

Final temp. = 23 °C

Change in temp ($\Delta T$ ) = (100 - 23 ) = 77 °C

Let specific heat capacity of aluminium be $s_A$.

The amount of heat lost by the aluminium sphere= $ m_1 s_A \Delta T=.047 \times 77 \times s_A$

Mass of water ($m_2$) = 0.25 kg

Mass of calorimeter ($m_3$) = 0.14 kg

Initial temp. of water and calorimeter = 20 °C

Final temp. of the mixture = 23 °C

Change in temp. ($\Delta T_2$) = 23 - 20 = 3 °C

The amount of heat gained by water and calorimeter = $m_2 s_w \Delta T_2 + m_3 s_c \Delta T_2$

$= (m_2s_w + m_3s_c) (\Delta T_2)$

$= (0.25 \times 4.18 \times 10^3 + 0.14 kg \times 0.386 \times 10^3) \times 3$

Now, $ \text {heat given by an aluminium sphere} =\text { heat absorbed by the water} + \\ \text {heat absorbed by copper calorimeter}$.

$.047 \times 77 \times s_A = (0.25 \times 4.18 \times 10^3 + 0.14 \times 0.386 \times 10^3 ) \times 3$

$s_A = 0.911 kJ kg^{-1} K^{-1}$

if .20 kg of hot tea at temperature 95° C is poured into a 150 g cup at temperature 25°. What will be equilibrium temperature of the tea and cup when steady state is reached .Assume no heat is lost to the surroundings?

Specific Heat of Tea= 4186 J/kg ° C

Specific Heat of cup= 840 J/kg ° C

Let T be the equilibrium temperature

Now

$\text {Heat lost by tea} =\text { heat gained by Cup}$

$4186 \times .20 \times (95 -T) = 840 \times .15 \times (T-25)$

T=86 ° C

**Notes****Assignments****Revision Notes**