Question

$10g$ of ice at $-{25}^{\circ }C$ is heated by a burner which is supplying heat energy at a rate of $250J/sec$. Calculate the time, in which the water formed from ice attains a temperature of ${90}^{\circ }C$. Specific heat capacity of ice is $2.1J{g^{-}}^{\circ }{C}^{-1}$ and specific latent heat of ice is $336J{g}^{-1}$.

Answer 1

Step 1: Given data

The quantity of ice is $10g$ and heat energy supplied at rate of $250J/sec$.

And the specific heat capacity of ice is $2.1J{g^{-}}^{\circ }{C}^{-1}$ and specific latent heat of ice is $336J{g}^{-1}$.

Step 2: Calculating the heat gained by ice to attain the temperature of $0°\mathrm{C}$

The heat required to raise the temperature of ice from $-{25}^{\circ }Cto{0}^{\circ }C$.

$$\begin{gathered} Heat=massofice\times sp.heatcapacityofice\times temperaturechange \\ \Rightarrow H_1=10\times 2.1\times (0-(-25\left)\right) \\ \Rightarrow H_1=525J \end{gathered}$$

Step 3: Calculating the Heat gained by ice to form water at $0°\mathrm{C}$

The heat required to Convert $0^{\circ }C$ ice to $0^{\circ }C$ water

$$\begin{gathered} H_2=massofwater\times latentheatofice \\ \Rightarrow H_2=10\times 336 \\ \Rightarrow H_2=3360J \end{gathered}$$

Step 4: Calculating the heat gained by water till ${90}^{\circ }C$

The heat required to convert the water $0^{\circ }Cto{90}^{\circ }C$

$$\begin{gathered} Heat=massofwater\times sp.heatofwater\times temperaturechange \\ \Rightarrow H_3=10\times 4.2\times 90 \\ \Rightarrow H_3=3780J \end{gathered}$$

Step 5: Calculating the time when water formed from ice

The total heat produced

$$\begin{gathered} =(525+3360+3780)J \\ =7665J \end{gathered}$$

Therefore the time required

$$\begin{gathered} =\frac{7665}{250}sec \\ =30.66sec \end{gathered}$$

Hence, the after the $30.66$ seconds water formed from ice and attains a temperature of ${90}^{\circ }C$.