Question

A 26 g ice cube at -15.0oC is placed in 204 g of water at 48.0oC. Find the final temperature of the system when equilibrium is reached. Ignore the heat capacity of the container and assume this is in a calorimeter, i.e. the system is thermally insulated from the surroundings. Give your answer in oC with 3 significant figures.

Specific heat of ice: 2.090 J/g K

Specific heat of water: 4.186 J/g K
Latent heat of fusion for water: 333 J/g

Answer 1

mass of ice,mi= 26 g

Temperature of ice,ti=-15.0^oC

mass of water,mw=204 g

Temperature of water,tw= 48.0^oC

Specific heat of ice: 2.090 J/g K

Specific heat of water: 4.186 J/g K

Latent heat of fusion for water: 333 J/g

Process (1): the heat required to change the temperature of 26 g = 0.026 kg ice is determined by the specific heat of the ice:

Q1=mi*ci*dt

The specific heat of ice is 2090 J/kg C, the initial temperature is (-15) and the final temperature is 0. Then the change of the temperature is 15:

Q1=mi*ci*dt=0.026*2090*15=815.1 J

Process (2): the heat required to melt the ice is determined by the specific latent heat of fusion:

Q2=mi*L_fusion

The specific latent heat of fusion is 333000 J/kg. Then

Q2=mi*L_fusion=0.026*333000=8658 J

Process (3): The ice now becomes the water and we need to increase the temperature of the water from 0 to the final temperature . The heat required to increase the temperature of the water is determined by the specific heat of the water (the mass of the water is equal to the mass of the ice):

Q3=mi*cw*dt

The specific heat of the water is 4186 J/kg C. Then

Q3=mi*cw*dt=0.026*4186*tf=109 tf

Then the total heat transferred from the ice is

Q1+Q2+Q3=815.1 J+8658+109tf=Q

(II) energy (heat) transferred to the system (2) has only one contribution: We just need to decrease the temperature of the water from the initial temperature 40 degrees to the final temperature . The mass of the water is .

mw=0.204 kg

The heat required to decrease the temperature of the water is determined by the specific heat of the water:

Qw=mct

The specific heat of the water is 4186 J/kg C, the change of the temperature of the water is,dt=(48-tf) . Then

Qw=0.204*4186*(48-tf)

Then the energy conservation takes the form:

9473+109tf=854(48-tf)

31516=963tf

tf=32.72^oC