Question

Calculate the quantity of heat required to convert $1.5kg$ of ice at $0{}^{\circ }C$ to water at ${{15}^{\circ }}C$ .

& {{L}_{ice}}=3.34\times {{10}^{5}}J.K{{g}^{-1}} \\\ & {{C}_{water}}=4180J.K{{g}^{-1}}.{}^{\circ }{{C}^{-1}} \\\ \end{aligned}$$

Answer 1

In the given question, there is a change in state from ice to water. When there is a change of state, the heat required for the given conversion would be the sum of the heat needed to raise the temperature of a particular state and the heat required to bring about changes in the state.

Complete step-by-step solution:
We have been given that:
The temperature of the ice is $0{}^{\circ }C$.
The temperature of the water is ${{15}^{\circ }}C$.
The mass of ice is $1.5kg$ .
The heat of fusion of ice and specific heat capacity of water is given respectively as:

& {{L}_{ice}}=3.34\times {{10}^{5}}J.K{{g}^{-1}} \\\ & {{C}_{water}}=4180J.K{{g}^{-1}}.{}^{\circ }{{C}^{-1}} \\\ \end{aligned}$$ Latent heat of fusion is defined as the heat required to convert a solid into a liquid or vapour, or a liquid into a vapour, without change of temperature. Specific heat capacity of a substance is defined as the amount of heat required to raise the temperature of $$1kg$$ a substance by $$1K$$ . The amount of heat required by material to change state at a given temperature is given by, $${{Q}_{s}}=m\times L$$ , where $$L$$ is the latent heat of fusion or latent heat of vaporization, depending on the state to which the material is changing, and $$m$$ is the mass of material. The amount of heat required by $$Mkg$$ of a substance to change its temperature is, $${{Q}_{t}}=M\times C\times \Delta T$$, where $$C$$ is the specific heat capacity of the material and $$\Delta T$$ is the temperature change. As per the given question, the ice is already at its melting temperature. The amount of heat needed by material to change its state at a given melting point is known as the heat of fusion. The heat of fusion is specific for different materials, in this case, the heat of fusion of ice is given as $$\begin{aligned} & {{L}_{ice}}=3.34\times {{10}^{5}}J.K{{g}^{-1}} \\\ & \\\ \end{aligned}$$ Thus, the amount of heat absorbed by ice to melt into the water at $$0{}^{\circ }C$$is $$\begin{aligned} & {{Q}_{s}}=m\times L \\\ & \Rightarrow {{Q}_{s}}=1.5\times 3.34\times {{10}^{5}}J \\\ & \Rightarrow {{Q}_{s}}=5.01\times {{10}^{5}}J \\\ \end{aligned}$$ Now, using the above amount of energy, the ice is converted into the water at $$0{}^{\circ }C$$. The amount of heat needed by water $$0{}^{\circ }C$$ to raise its temperature to $${{15}^{\circ }}C$$is, $$\begin{aligned} & {{Q}_{t}}=M\times C\times \Delta T \\\ & \Rightarrow {{Q}_{t}}=1.5\times 4180\times 15J \\\ & \Rightarrow {{Q}_{t}}=0.94\times {{10}^{5}}J \\\ \end{aligned}$$ Hence the total heat needed to convert $$1.5kg$$of ice at $$0{}^{\circ }C$$ to water at $${{15}^{\circ }}C$$ is $$\begin{aligned} & Q={{Q}_{s}}+{{Q}_{t}} \\\ & \Rightarrow Q=5.95\times {{10}^{5}}J \\\ \end{aligned}$$ **Note:-** The units of the physical quantities involved should be carefully monitored. Sometimes the latent heat and the specific heat might be given in calories per kilogram and they must be converted into joules per kilogram before proceeding with the calculations.