Question

A copper vessel of mass $100\,{\text{g}}$ contains $150\,{\text{g}}$ of water at $50\,^\circ {\text{C}}$ . How much ice is needed to cool it to $5\,^\circ {\text{C}}$ ?
Given: Specific heat capacity of copper $= 0.4\,{\text{J}}{{\text{g}}^{ - 1}}^\circ {{\text{C}}^{ - 1}}$
Specific heat capacity of water $= 4.2\,{\text{J}}{{\text{g}}^{ - 1}}^\circ {{\text{C}}^{ - 1}}$
Specific latent heat of fusion of ice $= 336\,{\text{J}}{{\text{g}}^{ - 1}}$

Answer 1

First of all, we will find the heat absorbed by the ice to melt along with the heat absorbed by it to reach the said temperature. Find the heat lost by the water and the copper vessel to lower the temperature. Use the idea that heat gained by ice is the heat lost by water and the vessel. Manipulate the equation and simplify accordingly to obtain the result.

Complete step by step answer:
In the given problem, mass of water is $150\,{\text{g}}$ .
Mass of the copper vessel is $100\,{\text{g}}$ .
Initial temperature of water and the vessel is $50\,^\circ {\text{C}}$ .
Final temperature of water and the vessel is $5\,^\circ {\text{C}}$ .
Specific heat capacity of copper $= 0.4\,{\text{J}}{{\text{g}}^{ - 1}}^\circ {{\text{C}}^{ - 1}}$
Specific heat capacity of water $= 4.2\,{\text{J}}{{\text{g}}^{ - 1}}^\circ {{\text{C}}^{ - 1}}$
Specific latent heat of fusion of ice $= 336\,{\text{J}}{{\text{g}}^{ - 1}}$

Before we begin, we will highlight the fact that ice is added to the water to lower its temperature. So, the question arises that where the heat will go, as we know, that heat is also a form of energy. According to the law of conservation of energy, energy can neither be destroyed nor be created, rather it can be converted to some other forms. Here, in this case the heat lost by the water and the vessel to lower its temperature is actually used to melt the ice which is added. This is the roadmap for the problem.

Total heat lost by the water and the vessel is given by the formula:
$H = {m_w}{c_w}\Delta t + {m_c}{c_c}\Delta t$ …… (1)
Where,
${m_w}$ indicates mass of water.
${c_w}$ indicates specific heat capacity of water.
${m_c}$ indicates mass of copper.
${c_c}$ indicates specific heat capacity of copper.
$\Delta t$ indicates temperature change.

Total heat gained by the ice to melt and elevate its temperature is given by the formula:
$H = {m_i}L + {m_i}{c_w}\delta t$ …… (2)
Where,
${m_i}$ indicates mass of ice.
$L$ indicates specific latent heat of ice fusion of ice.
${c_w}$ indicates specific heat capacity of water.
$\delta t$ indicates change in temperature.

Comparing equations (1) and (2), we get:
${m_w}{c_w}\Delta t + {m_c}{c_c}\Delta t = {m_i}L + {m_i}{c_w}\delta t$ …… (3)

Substituting the required values in the equation (3), we get:
$\left[ {150 \times 4.2 \times \left( {50 - 5} \right)} \right] + \left[ {100 \times 0.4 \times \left( {50 - 5} \right)} \right] = \left( {{m_i} \times 336} \right) + \left[ {{m_i} \times 4.2 \times \left( {5 - 0} \right)} \right] \\\ \Rightarrow \left( {150 \times 4.2 \times 45} \right) + \left( {100 \times 0.4 \times 45} \right) = \left( {{m_i} \times 336} \right) + \left( {{m_i} \times 4.2 \times 5} \right) \\\ \Rightarrow 28350 + 1800 = 336{m_i} + 21{m_i} \\\ \Rightarrow 30150 = 357{m_i} \\\$
Further simplifying, we get:
$30150 = 357{m_i} \\\ \Rightarrow {m_i} = \dfrac{{30150}}{{357}} \\\ \Rightarrow {m_i} = 84.5\,{\text{g}} \\\$

Hence, the required mass of ice is $84.5\,{\text{g}}$.

Note:
While solving this problem, always keep in mind that, magnitude of both the heat energies are equal, as the heat released by the water is used to melt the ice and lower the temperature. It is important to note that $\theta$ is the drop/fall/range of temperature, where the students tend to mess up this with a single reading of temperature. You can take the mass in kilograms also, that is not an issue, as the answer will remain the same.