Question

A 400 g ice cube at -20 ?C is placed in an aluminum cup whose initial temperature is 80 ?C . The system comes to an equilibrium temperature of 20 ?C .

What is the mass of the cup?

Express your answer with the appropriate units.

Answer 1

given

m_Ice = 400 g = 0.4 kg

T1 = -20 C

m_alu = ?

T2 = 80C

T = 20 C

we know,

C_Ice = 2108 J(kg C)

Lf = 3.33*10^5 J/kg

C_water = 4186 J/(kg C)

C_aluinum = 900 J/(kg C)

use Heat lost aluminum = heat gained by Ice

m_alu*C_alu*(T - T2) = m_Ice*C_Ice*(0 -(-20)) + m_Ice*Lf + m_Ice*C_water*(20 - 0 )

m_alu*900*(80 - 20) = 0.4*2108*20 + 0.4*3.33*10^5 + 0.4*4186*20

m_alu*900*60 = 183552

m_alu = 173552/(900*60)

= 3.21 kg or 3210 g <<<<<<<-----------Answer