# Heat transfer equation (temperature difference)

1. Nov 11, 2009

### MechaMZ

1. The problem statement, all variables and given/known data

For the equation mc$$\Delta$$T, i suppose $$\Delta$$T should be Tf - Ti.

However, at what condition we should let $$\Delta$$T be Ti - Tf?

I did come across a question which requires $$\Delta$$T be Ti - Tf for a heat lost by an object.

eg,
heat lost by copper = heat absorbed by beaker & water
mc(Ti - Tf) = mbeakercbeaker(Tf - Ti) + mwatercwater(Tf - Ti)

I think perhaps we need to let mc$$\Delta$$T as a positive value since it just a magnitude? however, the value is not the same(pretty close) if i let it be Ti - Tf.

Need help, thanks =)

2. Nov 17, 2009

### MechaMZ

anyone knows?

3. Nov 17, 2009

### Mapes

$mc\Delta T=mc(T_f-T_i)$ is always the energy gained. The energy lost by the copper was then $-mc\Delta T$, which is equivalent to the other expression.

4. Nov 17, 2009

### MechaMZ

however, for the example above, why the initial temp and final temp is reversed?

5. Nov 17, 2009

### Mapes

Because $-mc(T_f-T_i)=mc(T_i-T_f)$. Is this what you mean?

6. Nov 17, 2009

### MechaMZ

so we should write heat lost by copper as negative since the heat is released out, but the result of $-mc(T_f-T_i)$ will still be positive right?

7. Nov 17, 2009

### Mapes

If you're saying that we should include the minus sign prefactor because we're considering the amount of energy leaving, but that the sign of $-mc(T_f-T_i)$ is positive because $T_f<T_i$, I agree.