Question

work done in converting one gram of ice at -10 degree c into steam at 100 degree c is

Answer 1

Answer on Question #48485, Physics, Molecular Physics | Thermodynamics

Work done in converting one gram of ice at -10 degree C into steam at 100 degree C is

Solution:

The ice will warm up to 0 degrees Celsius, then melting and then warming to the final temperature 100 degrees Celsius, and steaming of water at 100 degrees Celsius.

Specific heat capacity, ice: $c_{\text{ice}} = 2.108 \, \text{kJ/kg} \cdot \text{K}$

Specific heat capacity, water: $c_{\text{water}} = 4.187 \, \text{kJ/kg} \cdot \text{K}$

The heat of fusion (or specific enthalpy of fusion) of ice is $L = 334 \, \text{kJ/kg}$

The latent heat of vaporization for water $H_v = 2260 \, \text{kJ/kg}$

$$m = 1 \times 10^{-3} \, \text{kg}$$

The energy to heat up the ice is the sum of the following

$$Q = c_{\text{ice}} m \Delta t_1 + L m + c_{\text{water}} m \Delta t_2 + H_v m$$ $$\Delta t_1 = 0 - (-10) = 10$$ $$\Delta t_2 = 100 - 0 = 100$$

Thus,

$$Q = 1 \cdot 10^{-3} (2.108 \cdot 10 + 334 + 4.187 \cdot 100 + 2260) \cdot 10^3 = 3033.78 \approx 3033.8 \, \text{J}$$

Answer: 3033.8 J

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