Heat transfer between two iron blocks

[ardour]

Homework Statement

Homework Equations

I'm not sure.

The Attempt at a Solution

I tried to solve this as you would with electric charges. I added up the temperatures and then divided by 2, to get (C) 17.5 degrees Celsius. The answer key gives the answer as (D) 20.0 degrees Celsius. I'm not sure what equations you would use to get this.

 

[kuruman]

Please try to figure out the relevant equations. This problem was not assigned to you out of the blue. It must have come at the end of a textbook chapter or lecture notes that surely contain some equations. Nevertheless, you can see why your answer is wrong because adding the temperatures and dividing by two will give the same final temperature regardless of the masses of the blocks. Does that sound reasonable? Hint: Yes, the final temperature is indeed some kind of average, but not the sum of temperatures divided by two.

 

[CWatters]

+1

It won't be the average temperature because the masses aren't the same. However don't think your approach is totally wrong.

 

[ardour]

Is q=mct the relevant equation? This question came from a practice test, so I'm not sure of the actual equation.

I tried setting mc(T-10) and 2mc(T-25) equal to each other, but got 40 for T. What am I doing wrong?

 

[kuruman]

The correct expression is $Q=mc~\Delta T$, where $\Delta T$ is the change in temperature, final minus initial. Why do you think the quantities you set equal are equal? What is physically going on here?

 

[ardour]

Q=mc ΔT

ΔT1= change in temperature for mass m
ΔT2= change in temperature for mass 2m

mcΔT1= 2mcΔT2
ΔT1=2ΔT2

The only answer which fits this is D, but I'm not sure how to calculate the temperature outright.

 

[kuruman]

mcΔT1=2mcΔT2

Can you explain in plain English and without symbols or equations what the two quantities that you are setting equal represent and why you think they are equal?

On Edit (Generous hint): Q stands for the heat gained by the mass. One mass gains heat heat, the other loses heat.