Question

You have a supply of ice at 0.0⁰Cand a glass containing 150gwater at 25⁰C.The enthalpy of fusion for ice is $\Delta H_{\text{fus}}=333{\text{Jg}}^{-1},$and the specific heat capacity of water is $4.18{\text{JK}}^{-1}{\text{g}}^{-1}.$How many grams of ice must be added to the glass (and melted) to reduce the temperature of the water to 0⁰C?

Short answer

The heat capacity of water is $-15.7\times {10}^3\text{J}$.

Ice must be added to reduce the temperature of the water to 0⁰C.

Step-by-step solution

Given data

The Mass of water, $\left(m\right)=150\text{g}$.

Initial temperature, $T_1={25}^{\circ }C$.

Final temperature, $T_2=0^{\circ }C$.

Enthalpy of fusion, $\Delta H_{\text{fus}}=333{\text{Jg}}^{-1}$.

Concept of the heat capacity

The ratio of heat absorbed by a substance to the temperature change is called heat capacity. In terms of the actual amount of material being considered, most frequently a mole, it is usually stated as calories per degree (the molecular weight in grams).

Calculation of heat capacity

The heat capacity can be determined by the formula which can be shown as: $q=m{c}_{\text{s}}\Delta T$ …. (i)

Where q is heat capacity, $\Delta T$is Change in temperature in K.

Now, put the value of given data in equation (i).

$$\begin{gathered} q=150g\times 4.18J{C}^{-1}{g}^{-1}\times (T_2-T_1) \\ =150g\times 4.18J{C}^{-1}{g}^{-1}\times (0^{\circ }C-25\circ C \\ =-15.7\times {10}^{3}J \end{gathered}$$

Calculation amount of ice

Enthalpy of fusion is $333{\text{Jg}}^{-1}$. The heat needed to fuse one gram of ice is 333J.

$15.7\times {10}^3$of heat will fuse.

Hence, the amount of ice will be:

$15.7\times {10}^3\text{J}\times \frac{1\text{g of ice}}{333\text{J}}=47.1\text{g}$

Therefore, the Heat capacity of water is $-15.7\times {10}^3\text{J}$.

47.1g ice must be added to reduce the temperature of the water to 0⁰ C.