# Homework Help: Solving specific heat problem with two materials

1. Feb 19, 2017

### HDTeach

1. The problem statement, all variables and given/known data
Imagine I have a 3.5 kg brick at 80 °C that I put in 10l of water at 20 °C. What will the final temperature of the water be?
SHC brick = 840 J kg-1 °C -1

2. Relevant equations
E = mc∆θ

3. The attempt at a solution

First I calculated the energy available from the brick to heat the water:
E = mc∆θ = 3.5*840*(80-20) = 176400J

Then I attempted to calculate the temperature change:
E = mc∆θ(for brick)+mc∆θ (for water)

176400 = 3.5*840*(80-∆θ) + 10*4200*(∆θ-20)

176400 = 2940(80-∆θ) + 42000(∆θ-20)

176400 – (2940*80) – (42000*-20) = (2940*-∆θ) + (42000*∆θ)

622440 = 39060*∆θ

∆θ = 16 degrees C

I am not sure this is correct though...any ideas? Have I over-complicated this?

2. Feb 19, 2017

### Staff: Mentor

Hi HDTeach,

Welcome to Physics Forums!

This would be the amount of heat that the brick gives up to bring its temperature down to 20C. Do you expect the brick to cool down all the way to the same value as that of the water's initial temperature?
The above equation shows both the water and brick giving up heat (they are being added together). Is that what happens when the materials come together? Or, does heat flow from one to the other?
Ask yourself if it makes sense that dropping a hot brick into room temperature water would cool the water (16 < 20).
No, you've just misunderstood how the heat is moving and so wrote an incorrect equation.

Heat will flow from the brick and to the water, cooling one and warning the other. The process stops when the temperature difference between the materials is zero, which will be at some temperature lying between their starting temperatures.

The trick is to write the ∆T's for each so that they incorporate the initial temperature and final (unknown) temperature for each material, and know that heat energy is conserved: One gains the same amount of heat as the other loses.

3. Feb 19, 2017

### HDTeach

Thank you! So would it be more like:

E = mc∆T = 3.5*840*(80-T) = 10*4200*(T-20)

235200 - 2940T = 42000T - 840000

1075200 = 44940T

T = 0.4 degrees C?

4. Feb 19, 2017

### Staff: Mentor

It looks good except for the final value. Redo your final calculation.

5. Feb 19, 2017

### HDTeach

Haha oops yes I see the error!

T = 23.9 degrees C

Thank you for your help. I had a worked example that confused me - is this correct in your view?

Question:
1.5 l of water from a kettle at 90°C is mixed with a bucket of cold water (10 l at 10 °C) to warm it up for washing a car.

Find the temperature of the mixed water, assuming no significant heat loss during the mixing. c of water = 4.2 kJ kg-1 °C -1

If there is no heat loss and no work done then the total energy of the system at the end is the same as at the start. Working with temperature differences from 10°C we have:

Initial energy = m c ΔT = 1.5 x 4200 x (90-10) [as 1.5 l has mass 1.5 kg] = 504 kJ

This is the amount of energy available to raise the temperature of all 11.5 l of water. Hence:

504 x 1000 = 11.5 x 4200 x (T-10) [where T is the final temperature]

T = 20.4°C

6. Feb 19, 2017

### Staff: Mentor

Yes. They took advantage of the fact that the two things being brought together were made of the same material and so had the same specific heat.

7. Feb 19, 2017

### HDTeach

Thank you, I just worked it out using the other method and got the same answer so it makes sense to me now! Thank you for all of your help :)

8. Feb 19, 2017

### Staff: Mentor

You're welcome!