Question

A slice of pizza has 500 kcal. If the pizza is burned and its heat is used to warm a 50-L container of cold water, what would be the approximate increase in the temperature of the water? (Note: A liter of cold water weighs about 1 kg).

Answer 1

Given:
The heat content of the pizza is  Q = 500 kcal 
The volume of the cold water is V = 50 L
The mass of a liter of cold water is 1 kg.

Calculate the mass, M,  of cold water.
M = (50 L)*(1 kg/L) = 50 kg 

The specific heat of water is
c = 4.184 kJ/(kg-K)

Also,
1 kcal = 4184 J, therefore
Q = (500 kcal)*(4184 J/kcal) = 2.092 x 10⁶ J

Let ΔT = increase in temperature of the cold water, °C (same as K).
Then
Q = M*c*ΔT
or 
ΔT = Q/(M*c)
$\Delta T = \frac{(2.092 \times 10^{6} J)}{(50 \, kg)(4.184 \times 10^{3} \, \frac{J}{kg-K)}) } =10 \,K=10 \,^{o}C$

Answer: The temperature increases by 10 °C (or 10 K)