# Heat Problem

1. ### someone1029

5
[SOLVED] Heat Problem

The question is:

The only equations known are Q=mcΔT, and Q=ml
Where Q= energy in joules. m=mass in kg, c= specific heat capacity [to 3 decimal places], ΔT= the change in temperature in either kelvin or Celsius, l= latent heat [also to 3 decimal places]

However I am not limited to these equations, just expected to use them. Use of other equations is fine, so long as they are within reason.

My train of thought was that, I need to find out and compare how many watts of heat will be absorbed by an identical body of mass over the same period of time at the two different temperatures, i just don't know how. I also think that this is one of those questions designed to trick, or at least make you really think. So I think that the short answer to the question is no, I just cant prove this right or wrong.

I have tried taking a theoretical 1kg aluminium block and putting them into each of the temperatures resulting in these equations:

8.90e+3*14=124600
8.90e+3*28=249200
as 249200 is 2*124600 then it is twice as hot

However I do not think this is right as these values are the amount of energy required to make this temperature change. Am I on the right track, or am I nowhere near the answer?

117

3. ### someone1029

5
No it isn't, -273°c is. Can you please explain the relevance of this though? I just chose 0°c as a point from which the blocks would get heated. If you convert the temperatures to kelvin you get exactly the same results.

4. ### edziura

117
14$$\circ$$C is 14 + 273 = 287 degrees above absolute zero. 28$$\circ$$C is 28 + 273 = 301 degrees above absolute zero.

Now $$\frac{301}{287}$$ = about 1.05, and so 28$$\circ$$C is about 5% hotter than 14$$\circ$$, not twice at hot.

A temperature of 2(287) - 273 = 301$$\circ$$C is twice as hot as 14$$\circ$$C.

The celsius scale is not an absolute T scale. The kelvin scale is. 0K is absobute zero and so, for example, 300K is twice as hot as 150K.

5. ### someone1029

5
Ok, I was just trying to over complicate things, as usual. Now that I look back at the problem I see how simple it is. Thanks for your help, no doubt I will call on it again some time in the future.