Question

Heat required to vaporize 4g of water by boiling at 373k is 2160 calories. The specific heat of water in this condition is

Answer 1

Hey dear,

◆ Answer-
22.572 kJ/kg°C

◆ Explanation-
# Given-
m = 4 g = 4×10^-3 kg
∆Ep = 2160 cal = 9028.8 J
∆T = 373-273 = 100 K

# Solution-
Enthalpy of vaporization is calculated by formula-
∆Ep = mc∆T
c = ∆Ep / m∆T
c = 9028.8 / (4×10^-3 × 100)
c = 22572 J/kg°C
c = 22.572 kJ/kg°C

Specific heat of water is 22.572 kJ/mol.

Hope this helps you..