Help with Water, Ice, and metal finding temperature

[raymondmorrow]

Homework Statement

A 50-g of ice, initially at 0.0C, is dropped into 200 g of water in an 80-g
aluminum container, both initially at 30C. What is the final equilibrium
temperature? (Specific heat for aluminum is 900 J/Kg*C, the specific heat of
water is 4186 J/Kg*C, and Lf=3.33 X10^5).

Homework Equations

The Attempt at a Solution

I am trying to help my wife with this. I have no clue what to do. Can you help me find somewhere it will show me what I need to do?

 

[Hootenanny]

Homework Statement

A 50-g of ice, initially at 0.0C, is dropped into 200 g of water in an 80-g
aluminum container, both initially at 30C. What is the final equilibrium
temperature? (Specific heat for aluminum is 900 J/Kg*C, the specific heat of
water is 4186 J/Kg*C, and Lf=3.33 X10^5).

Homework Equations

The Attempt at a Solution

I am trying to help my wife with this. I have no clue what to do. Can you help me find somewhere it will show me what I need to do?

Welcome to PF raymondmorrow,

What are your wife's thoughts on the question? What has she tried already?

 

[raymondmorrow]

heat lost by metal= M metal*C metal*change in temperature

heat gained by ice-water= (Q ice + Q water)+ m water*C water* change in temperature

heat lost by metal= heat gained by ice-water

I keep trying but I can't get the right answer (9.5 Celcius) Any help is very appreciated.

 

[Hootenanny]

heat gained by ice-water= (Q ice + Q water)+ m water*C water* change in temperature

Can you write this expression more explicitly? What are Q ice and Q water for example?

 

[raymondmorrow]

Heat lost by mutual:
(0.08kg)(900 j/kg*C)(30C-T)
=72 J/C(30C-T)

Ice to melt:
Q=(M ice)(lf)
=(0.05)(3.33x10^5)
=16650J
0.05kg * 0.2kg = 0.25kg

heat gained by water from ice:
(16650) + (0.25)(4.19 x 10 ^ 3)(T-0C)
16650+1047.5(T--0C)
16650+1047.5T

72 J/C(30C-T)=16650+1047.5T
2160-72J/CT=16650+1047.5T
-72 J/CT=14490+1047.5T (after subtracting 2160 from both sides)
-1195T=14490 (After subtracting 1047.5T from both sides)
=1.29

 

[hage567]

Heat lost by mutual:
(0.08kg)(900 j/kg*C)(30C-T)
=72 J/C(30C-T)

Ice to melt:
Q=(M ice)(lf)
=(0.05)(3.33x10^5)
=16650J
0.05kg * 0.2kg = 0.25kg

heat gained by water from ice:
(16650) + (0.25)(4.19 x 10 ^ 3)(T-0C)
16650+1047.5(T--0C)
16650+1047.5T

72 J/C(30C-T)=16650+1047.5T
2160-72J/CT=16650+1047.5T
-72 J/CT=14490+1047.5T (after subtracting 2160 from both sides)
-1195T=14490 (After subtracting 1047.5T from both sides)
=1.29

You can't add the masses of the two waters, since they are at different initial temperatures and the 50 g of water is taking heat away from the 200 g of water and the container. You need separate terms for each one. You should have 4 terms to work with:

the heat gained by the 50 g of ice at 0 degrees to become water at 0 degrees (Q1 let's say)
the heat gained by the 50 g of water now at 0 degrees to get to Tf (Q2)
the heat lost by the 200 g of water in the container to get to Tf (Q3)
the heat lost by the aluminum container to get to Tf (Q4)

Your equation should be laid out like

heat used to melt the ice (Q1) and warm the 50 g of water (Q2) = heat lost by 200 g of water (Q3) and heat lost by container (Q4).

Put the terms in the right place and solve for Tf.