Conceptual Physics

In the section on latent heat, we calculated the heat transferred to the same amount of ice as it melted to be 1.23×10^4 J. Here, we need the

heat transferred from the water, not to the ice. Since the water loses heat, we state this as í1.23×10^4 J, with a negative sign.

What is the strategy?

1. Calculate the temperature of the water after it loses heat to melt the ice, using the specific heat of water.


2. As the water reaches thermal equilibrium, heat transfers from the originally liquid water, to the melted ice. These are the only heat
   transfers in the system, so they sum to zero. Use this fact and the specific heat equation to calculate the temperature of the total mass
   of water at the end.

Physics principles and equations

The specific heat equation

Q = cmǻT

As the two masses of water reach thermal equilibrium, the heat transferred from the originally liquid water plus the heat transferred to the

melted ice must sum to zero.

QL + QS = 0

Step-by-step solution

First we calculate the temperature of the liquid water after it gives up heat to melt the ice. We use the specific heat of water, 4178 J/kg·K.

mass of liquid water mL = 0.160 kg

mass of solid ice mS = 0.0370 kg

heat transferred from water to melt ice Q = í1.23×10 (^4) J
initial temperature of liquid water TLi = 30.0°C
initial temperature of solid ice TSi = 0.00°C
temperature of liquid water after ice melts TLf
temperature of ice-melt TSf = 0.00°C
final temperature of liquid water plus melted ice T
heat transferred from liquid water for thermal equilibrium QL
heat transferred to melted ice for thermal equilibrium QS
specific heat of water c = 4178 J/kg·K

Step Reason

1. Q = cmǻT specific heat equation

2. substitute values

3. ǻT = –18.4 K = –18.4 C° solve

4. TLf = TLi + ǻT

TLf = 30.0°C + (–18.4°C)

TLf = 11.6°C

calculate water temperature

(^348) Copyright 2007 Kinetic Books Co. Chapter 18