Question

A calorimeter of mass 60g contains 180g of water at ${{29}^{\circ }}C$ . Find the final temperature of the mixture when 37.2g of ice at $-{{10}^{\circ }}C$ is added to it (specific heat capacity of water = \[4200\text{ }Jk{{g}^{-1}}{{K}^{-1}}\] , latent heat of ice = $336\times {{10}^{3}}\text{ Jk}{{\text{g}}^{-1}}$, specific capacity of ice =\[2100\text{ }Jk{{g}^{-1}}{{K}^{1}}\] , specific heat capacity of the calorimeter is $0.42J{{g}^{-1}}^{\circ }{{C}^{-1}}$ )

Answer 1

Hint: In the above question the water in the form of liquid and ice is added in a calorimeter. As a result a resultant or more specifically an equilibrium temperature is attained. Hence using the principle of calorimeter, the resultant temperature can be obtained accordingly.
Formula used:
${{Q}_{L}}=mL$
$\Delta Q=mC\Delta T$

Complete step-by-step solution:
The principle of calorimeter states that the heat gained by the cold body must be equal to that of the heat lost by the hot body. The heat i.e.$\Delta Q$ lost or gained by a body is given by,
$\Delta Q=mC\Delta T$
where m is the mass of the body,
C is the specific heat and $\Delta T$ is the resultant change in temperature.
If a body undergoes a change in state, it utilizes some heat called as the latent heat (${{Q}_{L}}$). This quantity is given by,
${{Q}_{L}}=mL$ where m is the mass of the body and L is the latent heat of fusion of the body during melting.
The calorimeter and the 180g of water are at the same temperature. Hence when ice is added in the calorimeter both the water and the calorimeter will lose heat. The heat will get utilized to change the temperature of the ice as well cause a change in state of the ice. Using this information we can imply,
$\Delta {{Q}_{i}}=\Delta {{Q}_{W}}+\Delta {{Q}_{C}}.....(1)$
Where $\Delta {{Q}_{i}}$ is the heat gained by ice,
$\Delta {{Q}_{W}}$ is the heat lost by water and $\Delta {{Q}_{C}}$ is the heat lost by a calorimeter. If ‘T’ is the final resultant temperature attained than,
$\begin{align}
  & \Delta {{Q}_{i}}={{Q}_{L}}+mC\Delta T \\
 & \Rightarrow \Delta {{Q}_{i}}=37.2g(336\times {{10}^{3}}\text{ Jk}{{\text{g}}^{-1}})+37.2g\times 2100\text{ }Jk{{g}^{-1}}{{K}^{1}}\times (T-(-10)) \\
 & \therefore \Delta {{Q}_{i}}=1.249\times {{10}^{10}}J+7.812\times {{10}^{7}}(T+10)J....(2) \\
\end{align}$
$\begin{align}
  & \Delta {{Q}_{W}}+\Delta {{Q}_{C}}=m{{C}_{W}}\Delta T+m{{C}_{C}}\Delta T \\
 & \Rightarrow \Delta {{Q}_{W}}+\Delta {{Q}_{C}}=180g\times 4200\text{ }Jk{{g}^{-1}}{{K}^{-1}}(29-T)+60g\times 0.42J{{g}^{-1}}^{\circ }{{C}^{-1}}\Delta T(29-T) \\
 & \therefore \Delta {{Q}_{W}}+\Delta {{Q}_{C}}=7.56\times {{10}^{8}}\text{ }(29-T)J+25.2(29-T)J....(3) \\
\end{align}$
From equation 1, 2 and 3 we get,
$\begin{align}
  & \Delta {{Q}_{i}}=\Delta {{Q}_{W}}+\Delta {{Q}_{C}} \\
 & 1.249\times {{10}^{10}}J+7.812\times {{10}^{7}}(T+10)J=7.56\times {{10}^{8}}\text{ }(29-T)J+25.2(29-T)J \\
 & \Rightarrow 1.249\times {{10}^{10}}+7.812\times {{10}^{7}}\times T+7.812\times {{10}^{8}}=7.56\times {{10}^{8}}\text{ }(29-T) \\
 & \Rightarrow 1.249\times {{10}^{10}}+7.812\times {{10}^{7}}\times T+7.812\times {{10}^{8}}=219.24\times {{10}^{8}}-7.56\times {{10}^{8}}\times T \\
 & \Rightarrow 7.812\times {{10}^{7}}\times T+75.6\times {{10}^{7}}\times T=2192.4\times {{10}^{7}}-78.12\times {{10}^{7}}-1249\times {{10}^{7}} \\
 & \Rightarrow 83.412\times T=865.28 \\
 & \therefore \text{T}={{10.37}^{\circ }}C\simeq {{10.4}^{\circ }}C \\
\end{align}$
Therefore the final temperature attained in equilibrium state is approximately equal to 10.4 degree Celsius.

Note: When the multiplier to a number is very big compared to the multiplier of a number which is to be added to it, that number can be ignored as per the laws of approximation. There is no heat loss to the surroundings. Hence the law of calorimetry holds valid in the above case. It is also to be noted that the difference on any temperature scale is always a constant.