Question

How much steam at $100^\circ {\rm{ C}}$ will just melt $2700{\rm{ g}}$ of ice at $- 10^\circ {\rm{ C}}$.(Sp. heat of ice = 0.5) and latent heat of steam $= 540{\rm{ cal }}{{\rm{g}}^{ - 1}}$).

Answer 1

We will be using the heat balance equation to calculate the amount of steam at $100^\circ {\rm{ C}}$ required to just melt the ice at $- 10^\circ {\rm{ C}}$. This heat balance equation can be obtained by writing the heat equation for steam and ice respectively and after that we will be equating these equations to calculate the mass of steam.

Complete step by step answer:

Given:
The specific heat of ice is ${\rm{C}} = 0.5$.
The latent heat of steam is ${L_1} = 540{\rm{ cal }}{{\rm{g}}^{ - 1}}$.
We have to find the amount of steam at $100^\circ {\rm{ C}}$ required to just melt $2700{\rm{ g}}$ of ice at $- 10^\circ {\rm{ C}}$.
We can easily understand that to just melt the ice we have to increase its temperature from $- 10^\circ {\rm{ C}}$ to $0^\circ {\rm{ C}}$ by supplying heat of the steam. In other words we can say that the ice is to be converted into water.
The amount of heat content of steam of mass ‘m’ is expressed as:
${Q_1} = m \times {L_1}$……(1)
Here L is the latent heat of steam which is defined as the amount of heat required by a body to change its phase at constant temperature.
The amount of heat required to raise the temperature of ice from $- 10^\circ {\rm{ C}}$ to $0^\circ {\rm{ C}}$ in order to convert it into water at $0^\circ {\rm{ C}}$.

{Q_2} = {m_2}CdT\\\ = {m_2}\left[ {C\left( {{T_2} - {T_1}} \right) + {L_2}} \right] \end{array}$$……(2) Here $${m_2}$$ is the mass of ice whose temperature is to be raised, $$dT$$ is the temperature difference and $${L_2}$$ is the latent heat of ice which is equal to 80 calorie per gram. Substitute $$2700{\rm{ g}}$$ for $${m_2}$$, $$0.5$$ for C, $$0^\circ {\rm{ C}}$$ for $${T_2}$$, $$80{\rm{ cal }}{{\rm{g}}^{ - 1}}$$ for $${L_2}$$and $$ - 10^\circ {\rm{ C}}$$ for $${T_1}$$ in equation (2). $${Q_2} = \left( {2700{\rm{ g}}} \right)\left( {0.5} \right)\left[ {0^\circ {\rm{ C}} - \left( { - 10^\circ {\rm{ C}}} \right)} \right] + \left[ {2700{\rm{ g}} \times 80{\rm{ cal }}{{\rm{g}}^{ - 1}}} \right]$$……(3) $${Q_2}$$ is the amount of heat required to raise the temperature of ice. Therefore equating the value of $${Q_2}$$ with the amount of heat obtained from equation (1). $$m \times {L_1} = \left( {2700{\rm{ g}}} \right)\left( {0.5} \right)\left[ {0^\circ {\rm{ C}} - \left( { - 10^\circ {\rm{ C}}} \right)} \right] + \left[ {2700{\rm{ g}} \times 80{\rm{ cal }}{{\rm{g}}^{ - 1}}} \right]$$ Substitute $$540{\rm{ cal }}{{\rm{g}}^{ - 1}}$$ for $${L_1}$$ in the above expression. $$\begin{array}{c} m \times \left( {540{\rm{ cal }}{{\rm{g}}^{ - 1}}} \right) = \left( {2700{\rm{ g}}} \right)\left( {0.5} \right)\left[ {0^\circ {\rm{ C}} - \left( { - 10^\circ {\rm{ C}}} \right)} \right] + \left[ {2700{\rm{ g}} \times 80{\rm{ cal }}{{\rm{g}}^{ - 1}}} \right]\\\ m = \dfrac{{229500}}{{540}}{\rm{ g}}\\\ m = {\rm{425 g}} \end{array}$$ The amount of steam at $$100^\circ {\rm{ C}}$$ required just to melt $$2700{\rm{ g}}$$ of ice at $$ - 10^\circ {\rm{ C}}$$ is $${\rm{425 g}}$$. **Note:** While writing the expression for the amount of heat required to just melt ice do not forget to add the latent heat required by the ice to get converted into water. It would be an added advantage if we remember the latent heat of water and ice to solve similar kinds of problems.