In a household radiator 1,000 g of steam at 100 degree C condenses, and the water cools to 90 degree C?

1. Use the latent heat of vaporization to find heat release during condensation:

Q_out_condns = m*h_v

2. Use the calorimetry formula, with the reverse of delta T:

Q_out_cool = m*c*(T1 - T2)

3. combine efforts:

Q_out_net = m*(h_v + c*(T1 - T2))

Data:

m:=1000 g; T1:=100 C; T2:=90 C; h_v:=540 cal/g; c:=1 cal/g-C;

note: calories are the thermochemical calories. NOT food Calories.

Results (prefix kilo added accordingly):

1. Q_out_condns = 540 kcal

2. Q_out_cool = 10 kcal

3. Q_out_net = 550 kcal

Household Radiator

whilst water at one hundred degree C is converted into steam at one hundred degree C, one extra warmth power of 540 energy per gram of water is utilised. as a result, the flexibility possesed with the help of steam is plenty greater than that of boiling water at one hundred degree C. this added power (that's stated as "LATENT warmth OF STEAM) is the reason for the severe burn.