Question

A calorimeter has mass $100\;{\rm{g}}$ and specific heat $0.1\;{{{\rm{kcal}}} {\left/ {\vphantom {{{\rm{kcal}}} {{\rm{kg}}{}^{\rm{o}}{\rm{C}}}}} \right. } {{\rm{kg}}{}^{\rm{o}}{\rm{C}}}}$. It contains $250\;{\rm{gm}}$of liquid at $30\;{}^{\rm{o}}{\rm{C}}$having specific heat of $0.4\;{{{\rm{kcal}}} {\left/ {\vphantom {{{\rm{kcal}}} {{\rm{kg}}{}^{\rm{o}}{\rm{C}}}}} \right. } {{\rm{kg}}{}^{\rm{o}}{\rm{C}}}}$, If we drop a piece of ice of mass $10\;{\rm{g}}$at ${0^ \circ }{\rm{C}}$, what will be the temperature of the mixture?

Answer 1

To solve this question, we will find the heat required by the ice at a particular temperature to convert into water at a particular temperature. Then we will find the heat released by the calorimeter and the liquid present in it. We will equate the heat released and that absorbed by the ice to get the final answer.

Complete step by step answer:
Given:
The mass of calorimeter is $M = 100\;{\rm{g}}$
The specific heat of calorimeter is $c = 0.1\;{{{\rm{kcal}}} {\left/ {\vphantom {{{\rm{kcal}}} {{\rm{kg}}{}^{\rm{o}}{\rm{C}}}}} \right. } {{\rm{kg}}{}^{\rm{o}}{\rm{C}}}}$.
The mass of liquid in the calorimeter is $m = 250\;{\rm{g}}$.
The specific heat of the liquid in calorimeter is ${c_l} = 0.4\;{{{\rm{kcal}}} {\left/ {\vphantom {{{\rm{kcal}}} {{\rm{kg}}{}^{\rm{o}}{\rm{C}}}}} \right. } {{\rm{kg}}{}^{\rm{o}}{\rm{C}}}}$.
The temperature of the liquid in a calorimeter is ${T_1} = 30\;{}^{\rm{o}}{\rm{C}}$.
The mass of ice is ${m_i} = 10\;{\rm{g}}$
We will assume the final temperature of the mixture as ${T_2}\,{}^{\rm{o}}{\rm{C}}$.
We know that the latent heat required to convert ice to water is $334{{\,{\rm{J}}} {\left/ {\vphantom {{\,{\rm{J}}} {{\rm{kg}}}}} \right. } {{\rm{kg}}}}$that is $80{{\;{\rm{cal}}} {\left/ {\vphantom {{\;{\rm{cal}}} {\rm{g}}}} \right. } {\rm{g}}}$.
We will write an expression for the heat required to convert $10\;{\rm{g}}$ice at ${0^ \circ }{\rm{C}}$to water at $0\,{}^{\rm{o}}{\rm{C}}$.
${Q_1} = {m_i} \times L.H.$
We will substitute $10\;{\rm{g}}$for ${m_i}$and $80\;{{{\rm{cal}}} {\left/ {\vphantom {{{\rm{cal}}} {\rm{g}}}} \right. } {\rm{g}}}$for $L.H.$in the above expression.

{Q_1} = 10\;{\rm{g}} \times 80{{\;{\rm{cal}}} {\left/ {\vphantom {{\;{\rm{cal}}} {\rm{g}}}} \right. } {\rm{g}}}\\\ {Q_1} = 800\;\;{\rm{cal}} \end{array}$$ We will write an expression for the heat required to convert $10\;{\rm{g}}$water at ${0^ \circ }{\rm{C}}$to water at ${T_2}\,{}^{\rm{o}}{\rm{C}}$. ${Q_2} = {m_w} \times {c_w} \times \left( {{T_2} - {T_1}} \right)$ We will substitute $10\;{\rm{g}}$for ${m_i}$and we know that specific heat of water is $1\;{{{\rm{cal}}} {\left/ {\vphantom {{{\rm{cal}}} {\rm{g}}}} \right. } {\rm{g}}}$. $$\begin{array}{l} {Q_2} = 10\;{\rm{g}} \times 1{{\;{\rm{cal}}} {\left/ {\vphantom {{\;{\rm{cal}}} {{\rm{kg}} \times \left( {{T_2} - 0} \right)}}} \right. } {{\rm{kg}} \times \left( {{T_2} - 0} \right)}}\\\ {Q_2} = 10{T_2}\;{\rm{cal}} \end{array}$$ We can write the total energy required to convert$10\;{\rm{g}}$ice at ${0^ \circ }{\rm{C}}$to water at ${T_2}\,{}^{\rm{o}}{\rm{C}}$. $Q = {Q_1} + {Q_2}$ We will substitute ${Q_1}$and ${Q_2}$in the above expression we will get: $\begin{array}{l} Q = 800\;\;{\rm{cal}} + 10 \times {T_2}\;{\rm{cal}}\\\ Q = \left( {800 + 10{T_2}} \right)\;{\rm{cal}} \end{array}$ Now, we will write the expression for heat released in order to change the temperature of calorimeter from $30\;{}^{\rm{o}}{\rm{C}}$to ${T_2}\;{}^{\rm{o}}{\rm{C}}$ ${Q_c} = M \times c \times \left( {{T_2} - {T_1}} \right)$ We will substitute $100\;{\rm{g}}$for $M$, $$0.1\;{{{\rm{kcal}}} {\left/ {\vphantom {{{\rm{kcal}}} {{\rm{kg}}{}^{\rm{o}}{\rm{C}}}}} \right. } {{\rm{kg}}{}^{\rm{o}}{\rm{C}}}}$$for $c$and $30\;{}^{\rm{o}}{\rm{C}}$for ${T_2}$in the above expression. $\begin{array}{l} {Q_c} = 100\;{\rm{g}} \times \dfrac{{{{10}^{ - 3}}\;{\rm{kg}}}}{{1\;{\rm{g}}}} \times 0.1\;{{{\rm{kcal}}} {\left/ {\vphantom {{{\rm{kcal}}} {{\rm{kg}}{}^{\rm{o}}{\rm{C}}}}} \right. } {{\rm{kg}}{}^{\rm{o}}{\rm{C}}}} \times \left( {{{30}^ \circ }C - {T_2}} \right)\\\ {Q_c} = {10^{ - 2}} \times \left( {{{30}^ \circ }C - {T_2}} \right)\;{\rm{kcal}} \end{array}$ We know that heat required to change the temperature of liquid in the calorimeter will be $${Q_l} = m \times {c_l} \times \left( {{T_2} - {T_1}} \right)$$ We will substitute $250\;{\rm{g}}$for $m$, $$0.4\;{{{\rm{kcal}}} {\left/ {\vphantom {{{\rm{kcal}}} {{\rm{kg}}{}^{\rm{o}}{\rm{C}}}}} \right. } {{\rm{kg}}{}^{\rm{o}}{\rm{C}}}}$$for $c$and $30\;{}^{\rm{o}}{\rm{C}}$for ${T_2}$in the above expression. $$\begin{array}{l} {Q_l} = 250\;{\rm{g}} \times \dfrac{{{{10}^{ - 3}}\;{\rm{kg}}}}{{1\;{\rm{g}}}} \times 0.4\;{{{\rm{kcal}}} {\left/ {\vphantom {{{\rm{kcal}}} {{\rm{kg}}{}^{\rm{o}}{\rm{C}}}}} \right. } {{\rm{kg}}{}^{\rm{o}}{\rm{C}}}} \times \left( {{{30}^ \circ }C - {T_2}} \right)\\\ {Q_l} = 10 \times {10^{ - 2}}\left( {{{30}^ \circ }C - {T_2}} \right)\;{\rm{kcal}} \end{array}$$ Now we can write the expression for the total heat released in the above process. ${Q_{total}} = {Q_c} + {Q_l}$ We will substitute the values of and ${Q_l}$in the above expression. $\begin{array}{l} {Q_{total}} = {10^{ - 2}} \times \left( {{{30}^ \circ }C - {T_2}} \right)\;{\rm{kcal}} + 10 \times {10^{ - 2}}\left( {{{30}^ \circ }C - {T_2}} \right)\;{\rm{kcal}}\\\ {{\rm{Q}}_{total}} = 11 \times {10^{ - 2}}\left( {{{30}^ \circ }C - {T_2}} \right)\;{\rm{kcal}} \end{array}$ We know that thus heat released will be absorbed by the ice. Hence, we will equate the ${Q_{total}}$with $Q$. This can be expressed as: $\begin{array}{l} \left( {11 \times {{10}^{ - 2}}\left( {{{30}^ \circ }C - {T_2}} \right)\;{\rm{kcal}}} \right) \times \dfrac{{{\rm{1}}{{\rm{0}}^{\rm{3}}}\;{\rm{cal}}}}{{{\rm{1}}\;{\rm{kcal}}}} = \left( {800 + {T_2}} \right)\;{\rm{cal}}\\\ {\rm{800}} + 10{T_2} = 110\left( {{{30}^ \circ }C - {T_2}} \right)\\\ {T_2} = 20.83\;{}^{\rm{o}}{\rm{C}} \end{array}$ Therefore, the temperature of the mixture will be ${\rm{20}}{\rm{.83}}\;{}^{\rm{o}}{\rm{C}}$. **Note:** An essential thing to consider in this question is the energy balance. We know that heat released by the calorimeter will be equal to the heat absorbed by the ice. By comparing both the energies, we can find the final temperature of the mixture.