Calorimetry

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 (s_w) = $4.18 \times 10^3 J kg^{-1} K^{-1}$
Specific heat capacity of copper calorimeter (s_c)= $0.386 \times 10^3 J kg^{-1} K^{-1}$
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