Thermodynamics - Hot copper block dropped in ice water

[nighfallraid]

Homework Statement

A 0.20 kg block of copper is heated in a laboratory oven to a temperature of 180o
C. It is added to an insulated
container of ice and water which is at a temperature of 0o
C. The mass of ice is 0.10 kg and the mass of water
is 0.90 kg.
i) Calculate the heat required to melt all the ice in the container.
ii) Calculate the heat released if the 0.20 kg block of copper is cooled from 180o
C to 0o
C.
iii) Find the final temperature of the contents of the container.
Neglect the heat capacity of the container. Assume no heat is lost to the surroundings.

Relevant Equations

The specific heat capacity of copper is 381 J kg -1 K-1
The specific heat capacity of liquid water is 4.19 kJ kg -1 K-1
The latent heat of fusion of water is 333 kJ kg-1
The latent heat of vaporisation of water is 2260 kJ kg-1

1) Q=m*L
-.10*333
2)Q=mc deltat
-.10*381*180-0
im unsure about the rest

 

[erobz]

I think as a first step to part (iii) you want to figure out what temperature the copper would be if it melted all the ice.

 

[Chestermiller]

What are the units of the answers in parts 1 and 2

 

[nighfallraid]

What are the units of the answers in parts 1 and 2

What are the units of the answers in parts 1 and 2

kJ

 

[nighfallraid]

I think as a first step to part (iii) you want to figure out what temperature the copper would be if it melted all the ice.

can you show me how ?

 

[haruspex]

2)Q=mc deltat
-.10*381*180-0

What is the mass of the copper?
Also it will release heat, so "heat released" should be positive.

1) Q=m*L
-.10*333

Heat will need to be added to melt the ice, so "heat required" is positive.

For iii, you do not know at the start whether all the ice will melt. Just assume whatever seems the likelier of the two possibilities and calculate based on that.
If you get a silly answer (like, negative mass of ice left), switch to the other.

 

[erobz]

can you show me how ?

You know the heat required to be absorbed by the ice to melt it $Q_{melt}$, you calculated it in (i). Equate that to the change in internal energy of the copper ball:

$$ \Delta U_c = - m_{ice}L $$

 

[erobz]

kJ

the units in (ii) are not $\rm{kJ}$.

 

[erobz]

What parts (i) and (ii) are begging at is for you to use them to determine if the final state has ice or not. In order for all the ice to be melted in the final state of equilibrium you have a certain amount of heat that needs to be absorbed by the ice before the supplier of said heat (the copper mass) reaches the temperature of the ice ( i.e. $0~{}^{\circ}C$ )., because heat will not be transferred without thermal gradient. Can the copper mass supply enough heat to do that?

 

[Chestermiller]

kJ

The specific heat capacity of copper is 381 J kg -1 K-1