Question

Ice at –10°C is to be converted into steam at 110°C. Mass of ice is \(10^{–3}\) kg. What amount of heat is required?

(1) \(\Delta Q = 730\ cal\) 

(2) \(\Delta Q = 900\ cal\)

(3) \( \Delta Q =  1210  \ cal \) 

(4) \(\Delta Q = 870\ cal\)

Answer 1

Correct option is : (1) \(\Delta Q = 730 \ cal\)

–10°C ice to 0°C ice \(\rightarrow\) 0°C ice to 0°C water + 0°C water to 100°C water + 100°C water to 100°C steam + 110°C steam.