If you start with 163 g of water at 29◦C, how much heat must you add to convert all the liquid into vapor at 100◦C? Assume no heat is lost to the surroundings.
Answer in units of kJ.

48.37514 kj

Explanation:

Given data:

Mass of water = 163 g

Initial temperature = 29°C

Final temperature = 100°C

Heat added = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Specific heat capacity of water is 4.18 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 = Final temperature - initial temperature

ΔT =  100°C - 29°C

ΔT =  71°C

Q = 163 g × 4.18 j/g.°C × 71°C

Q = 48375.14 j

Joule to Kj conversion:

48375.14 /1000 = 48.37514 kj