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 = ((2.092 * 10^(6) J))/((50 \, kg)(4.184 * 10^(3) \, (J)/(kg-K))) ) =10 \,K=10 \,^(o)C`

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