Question

A 4.00 - kg silver ingot is taken from furnace, where its temperature is $\mathbf{750}\mathbf{.}\mathbf{0}\mathbf{°}\mathbf{C}$,and placed on a large block of ice at$\mathbf{0}\mathbf{.}\mathbf{0}\mathbf{°}\mathbf{C}$.Assuming that all the heat given up by the silver is used to melt the ice, how much ice is melted?

Short answer

The mass of melted ice is 2.10kg

Given: The mass of silver is ${\mathrm{M}}_{\mathrm{Ag}}=4.00\mathrm{kg}$

The temperature of silver ingot is ${\mathrm{T}}_1=750.0°\mathrm{C}$

The temperature of ice is ${\mathrm{T}}_2=0.0°\mathrm{C}$

Step-by-step solution

The conservation of heat energy

The heat gain by gain of ice is equals to the heat lost by silver ingot

${\mathrm{Q}}_{\mathrm{Ice}}=-{\mathrm{Q}}_{\mathrm{Ag}}$

Where, ${\mathrm{Q}}_{\mathrm{Ice}}$is the heat gain ice and ${\mathrm{Q}}_{\mathrm{Ag}}$is the heat gain by silver ingot.

Heat gain by ice and heat lost by silver ingot

Heat gain by ice for melting is given by

${\mathrm{Q}}_{\mathrm{ice}}={\mathrm{M}}_{\mathrm{ice}}{\mathrm{L}}_{\mathrm{f}}$

Where, ${\mathrm{M}}_{\mathrm{ice}}$mass of melted ice and the latent heat of fusion of ice is

${\mathrm{L}}_{\mathrm{f}}=33.4\times {10}^3\mathrm{J}/\mathrm{kg}$

Heat lost by silver ingot

${\mathrm{Q}}_{\mathrm{Ag}}={\mathrm{M}}_{\mathrm{Ag}}{\mathrm{C}}_{\mathrm{Ag}}∆\mathrm{T}$

Where, ${\mathrm{M}}_{\mathrm{Ag}}$mass of silver ingot and the specific heat capacity of silver ingot is${\mathrm{C}}_{\mathrm{Ag}}=234\mathrm{J}/\mathrm{kg}.\mathrm{K}$.

Calculation of mass of ice

Using equation

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{ice}}=-{\mathrm{Q}}_{\mathrm{Ag}} \\ {\mathrm{M}}_{\mathrm{ice}}=\frac{-{\mathrm{M}}_{\mathrm{Ag}}{\mathrm{C}}_{\mathrm{Ag}}∆\mathrm{T}}{{\mathrm{L}}_{\mathrm{f}}} \end{gathered}$$

Now, putting the values of constants in above equation

$$\begin{gathered} {\mathrm{M}}_{\mathrm{ice}}=\frac{-\left(4\mathrm{kg}\times 243\mathrm{J}/\mathrm{kg}.\mathrm{K}\times \left(0.0°\mathrm{K}-75.0°\mathrm{K}\right)\right)}{33.4\times {10}^3\mathrm{J}/\mathrm{kg}} \\ {\mathrm{M}}_{\mathrm{ice}}=2.10\mathrm{kg} \end{gathered}$$

Thus, The mass of melted ice is 2.10kg