Question

work done in converting one gram of ice at minus 10 degree Celsius to steam at hundred degree celsius is

Answer 1

Q = mass * Specific heat * ∆ T 

For ice, Q = 1 * 2.09 * 10 = 20.9 J 
For water, Q = 1 * 4.186 * 100 = 418.6 J 


For melting the ice, use the following equation. 

Q = mass * Heat of fusion 
Heat of fusion 333 J/g 
Q = 1 * 333 = 333 J 

For boiling the water, use the following equation. 

Q = mass * Heat of vaporization 
Heat of vaporization = 2260 J/g 
Q = 1 * 2260 = 2260 J 

Total = 20.9 + 418.6 + 333 + 2260 = 3032.5 J = 3.0325 kg 

Answer 2

Answer:

This is a very simple one please learn the values of specific heat and latent heats

Explanation: