Calculate the amount of heat in kilojoules required when 13.0 g of water at 52.5°C is converted to steam at 238.2°C.

Q = 10.10059 KJ

Explanation:

Given data:

Amount of water = 13.0 g

Initial temperature = 52.5°C

Final temperature = 238.2°C

Heat required to convert it into steam = ?

Solution:

Specific heat capacity of water is 4.184 J/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 238.2°C - 52.5°C

ΔT = 185.7 °C

by putting values,

Q = 13.0 g × 4.184 J/g.°C × 185.7 °C

Q = 10100.59 J

Jolue to Kj:

10100.59j × 1 KJ/1000 J

Q = 10.10059 KJ