Question

(8%) Problem 7: A 0.25-kg aluminum bowl holding 0.75 kg of soup at 25.0°C is placed in a freezer. show answer Incorrect Answer What is the final temperature, in degrees Celsius, if 377 kJ of energy is transferred from the bowl and soup, assuming the soup’s thermal properties are the same as that of water? Specifically, you can assume the latent heat of fusion for the soup is 334 kJ/kg.

Specific heat (c) kcal/kg.C 0.215 0.20 Substances Solids Aluminum J/kg.C 900 Concrete 840 Copper 387 0.0924 Glass 840 0.20 Ice (average) Liquids 2090 0.50 Water 4186 1.000 Gases 1520 (2020) 0.363(0.482) Steam (100°C) Otheexpertta.com

Answer 1

here,

mass of alumunium , m1 = 0.25 kg

mass of soup , m2 = 0.75 degree

initial temperature , Ti = 25 degree C

let the final temperature be Tf

the heat energy transferred for Tf = 0 degree C, Q1 = m1 * Ca * ( Ti - Tf) + m2 * Cw * ( Ti - Tf)

Q1 = 0.25 * 900 * ( 25 - 0) + 0.75 * 4186 * ( 25 - 0) J

Q1 = 84112.5 J

so, the soup must be freezed

let the final temperature be Tf

equating the energy

Q1 + m1 * Ca * ( 0 - Tf) + m2 * Lf + m2 * Ci * ( 0 - Tf) = 377 * 10^3

84112.5 + 0.25 * 900 * ( 0 - Tf) + 0.75 * 334000 + 0.75 * 2100 * ( 0 - Tf) = 377 * 10^3

solving for Tf

Tf = - 23.5 degree C

the final temperature is - 23.5 degree C