Question

Heat energy is supplied at a constant rate to 100 g of ice at $0^{\circ }\mathrm{C}$.The ice is converted into the water at $0^{\circ }\mathrm{C}$.The ice is converted into water at $0^{\circ }\mathrm{C}$ in 2 min. How much time will be required to raise the temperature of the water from $0^{\circ }\mathrm{C}$ to ${20}^{\circ }\mathrm{C}$?

(Take, specific heat capacity of water = $4.2{\mathrm{Jg}}^{-1}{}^{\circ }\mathrm{C}^{-1}$ and latent heat of ice = $336\mathrm{J}/\mathrm{g}$)

Answer 1

Step 1: Given data

Mass $\mathrm{m}=100\mathrm{g}$

Time $\mathrm{t}=2\mathrm{minutes}=2\times 60=120\mathrm{s}$

Specific heat capacity of water $c=4.2{\mathrm{Jg}}^{-1}{}^{\circ }\mathrm{C}^{-1}$

Latent heat of ice $\mathrm{L}=336\mathrm{J}/\mathrm{g}$

Change in temperature $∆\mathrm{T}=\left({20}^{\circ }\mathrm{C}-0^{\circ }\mathrm{C}\right)={20}^{\circ }\mathrm{C}$

Step 2: Calculation of the time required to raise the temperature of the water from $0^{\circ }\mathrm{C}$ to ${20}^{\circ }\mathrm{C}$

$$\begin{gathered} \mathrm{Heat}\mathrm{energy}=\mathrm{Power}\times \mathrm{time} \\ Q=\mathrm{P}\times \mathrm{t}\text{'} \end{gathered}$$

Heat energy $Q=mc∆\mathrm{T}$

On substitution:

$$\begin{gathered} mc∆\mathrm{T}=\frac{\mathrm{mL}}{\mathrm{t}}\times \mathrm{t}\text{'} \\ 100\mathrm{g}\times 4.2\mathrm{J}.{\mathrm{g}}^{-1}.{}^{\circ }\mathrm{C}^{-1}\times {20}^{\circ }\mathrm{C}=\frac{336\mathrm{J}/\mathrm{g}\times 100\mathrm{g}}{120\mathrm{s}}\times \mathrm{t}\text{'} \\ 8400=280\times \mathrm{t}\text{'} \\ \mathrm{t}\text{'}=\frac{8400}{280} \\ \mathrm{t}\text{'}=30\sec \end{gathered}$$

Thus the time required to raise the temperature of the water from $0^{\circ }\mathrm{C}$ to ${20}^{\circ }\mathrm{C}$ is 30 sec.