Question

1. A 0.400 kg sample of aluminum (c = 9.20 x 102 J/kg°C) at 95.0 °C is dropped into a 0.550 kg pot of water which is at 18.0 °C. What temperature will the mixture come to?
2. How much thermal energy is absorbed by a 0.0220 kg ice cube as it melts? The latent heat of fusion of water is 3.40x 10^5 J/kg.

Answer 1

Equation: Qlost + Qgained = 0
m1 c1∆T1+m2 c2 ∆T2=0
Substitute:
(0.550kg)(4200 J/kg℃)(T2-18.0 ℃)+(0.400 kg)(920 J/kg℃)(T2-95.0 ℃)=0
2310(T2-18.0 ℃)+368(T2-95.0 ℃)=0
2310T2-41580+368T2-34960=0
2678T2-76540=0
2678T2=76540
T2=28℃
Statement: The mixture will come to an equilibrium temperature of 28℃.

2.
Given: m=0.0220 kg,Lf=3.40x10^5 J/kg
Unknown: Q=?
Equation: Q=mLf
Substitute: Q=(0.0220 kg)(3.40x10^5 J/kg)
Q=7480
Q=7480 J
Statement: Therefore, 7.5x10^3 J worth of thermal energy is absorbed by the ice as it melts.

Answer 2

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