Question

A 0.250-kg aluminum bowl holding 0.800 kg of soup at 26.2°C is placed in a freezer. What is the final temperature if 428 kJ of energy is transferred from the bowl and soup? Assume the soup has the same thermal properties as that of water, the specific heat of the liquid soup is 1.00 kcal/(kg · °C), frozen soup is 0.500 kcal/(kg · °C), and the latent heat of fusion is 79.8 kcal/kg. The specific heat of aluminum is 0.215 kcal/(kg · °C).
How much heat is needed to bring the soup and bowl to 0°C? How much heat is needed to freeze the soup? How much heat is needed to lower the temperature of the soup and bowl to below freezing?°C

Answer 1

here,

mass of bowl , m1 = 0.25 kg

mass of soup , m2 = 0.8 kg

let the final temprature be Tf

heat energy lost = m1 * Ca * ( 26.2 - Tf) + m2 * Cw * (26.2 - 0) + m2 * Lf + m2 * Ci * ( 0 - TF)

428000 = 0.25 * 900 * ( 26.2 - Tf) + 0.8 * 4186 * (26.2) + 0.8 * 334000 + 0.8 * 2150 * ( 0 - Tf)

solving for Tf

Tf = - 34.5 degree

the final temprature is - 34.5 degree C

the heat required to bring the soup and bowl to 0 degree C , Q1 = m1 * Ca * ( 26.2 - 0) + m2 * Cw * (26.2 - 0)

Q1 = 0.25 * 900 * ( 26.2 - 0) + 0.8 * 4186 * (26.2) = 93633.6 J

the heat required to freeze the soup , Q2 = m2 * LF

Q2 = 267200 J

the heat needed to lower the temperature of the soup and bowl to below freezing , Q3 = 428000 - 93633.3 - 267200 J

Q3 = 67166.7 J