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.
$\text{Heat Lost by hot objects} =\text {Heat Gained by cold object}$
Solved examples
Question 1
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 (sw) = $4.18 \times 10^3 J kg^{-1} K^{-1}$
Specific heat capacity of copper calorimeter (sc)= $0.386 \times 10^3 J kg^{-1} K^{-1}$ Solution
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}$.
$.047 \times 77 \times s_A = (0.25 \times 4.18 \times 10^3 + 0.14 \times 0.386 \times 10^3 ) \times 3$
$s_A = 0.911 kJ kg^{-1} K^{-1}$
Question 2
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 Solution
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
Watch this tutorial for more information on Heat And Calorimetry
