Work done in converting one gram of ice at – 10°C into steam at 100°C is

The correct option is 3045 J.

To find the total work done (energy required) to convert 1 gram of ice at -10°C to steam at 100°C, we need to calculate the heat energy required at each phase change and temperature change stage.

Let the mass of ice be:
$m=1\text{g}$

We will use the following standard thermal constants:
- Specific heat of ice ($c_{\text{ice}}$) = 0.5 cal/g°C
- Latent heat of fusion of ice ($L_f$) = 80 cal/g
- Specific heat of water ($c_{\text{water}}$) = 1 cal/g°C
- Latent heat of vaporization of water ($L_v$) = 540 cal/g
- Mechanical equivalent of heat ($J$) = 4.2 J/cal

Step 1: Heat required to raise the temperature of ice from -10°C to 0°C ($Q_1$)
$Q_1=m\cdot c_{\text{ice}}\cdot \Delta T$
$Q_1=1\cdot 0.5\cdot (0-(-10\left)\right)=5\text{cal}$

Step 2: Heat required to melt ice at 0°C into water at 0°C ($Q_2$)
$Q_2=m\cdot L_f$
$Q_2=1\cdot 80=80\text{cal}$

Step 3: Heat required to raise the temperature of water from 0°C to 100°C ($Q_3$)
$Q_3=m\cdot c_{\text{water}}\cdot \Delta T$
$Q_3=1\cdot 1\cdot (100-0)=100\text{cal}$

Step 4: Heat required to convert water at 100°C into steam at 100°C ($Q_4$)
$Q_4=m\cdot L_v$
$Q_4=1\cdot 540=540\text{cal}$

Step 5: Total heat energy required ($Q$)
$Q=Q_1+Q_2+Q_3+Q_4$
$Q=5+80+100+540=725\text{cal}$

Step 6: Conversion of heat energy to work (Joules)
Using the relation $W=J\cdot Q$:
$W=4.2\cdot 725=3045\text{J}$

Thus, the total work done to convert 1 gram of ice at -10°C to steam at 100°C is 3045 J.