Heat dissipated by a steel disk

[Bolter]

Homework Statement

See below

Relevant Equations

Q = mc(deta T)

Hi everyone

This is a quick Q but I don't understand why I got it wrong

This is what I have done

Could anyone let me know where I went wrong? Thanks for any help!

 

[etotheipi]

You're just getting muddled with units. $$T_2 - T_1 = \frac{-99\text{kJ}}{mc} \approx -139 K$$Remembering that a temperature difference expressed in Kelvin is equivalent to a temperature difference expressed in Celsius, you could have $$T_1 \approx ( 290 + 139) \text{K}$$or$$T_1 \approx (17 + 139)^o \text{C}$$Temperature is a bit funny when it comes to units, so you need to be careful!

 

[Chestermiller]

Homework Statement:: See below
Relevant Equations:: Q = mc(deta T)

Hi everyone

This is a quick Q but I don't understand why I got it wrong

View attachment 271308

This is what I have doneView attachment 271309

Could anyone let me know where I went wrong? Thanks for any help!

This problem statement makes no sense to me. What idiot dreamt this up?

 

[etotheipi]

This problem statement makes no sense to me. What idiot dreamt this up?

Yeah, it's a very poor question. I suspect what they were trying to describe was a process in which all the heat dissipated during the braking goes into raising the temperature of the steel disk to some temperature $T$, and then the disk is allowed to cool whilst in thermal contact with the constant temperature surroundings to a final temperature which approaches the temperature of the surroundings. Definitely quite an artificial setup, and not very physical!

 

[Bolter]

You're just getting muddled with units. $$T_2 - T_1 = \frac{-99\text{kJ}}{mc} \approx -139 K$$Remembering that a temperature difference expressed in Kelvin is equivalent to a temperature difference expressed in Celsius, you could have $$T_1 \approx ( 290 + 139) \text{K}$$or$$T_1 \approx (17 + 139)^o \text{C}$$Temperature is a bit funny when it comes to units, so you need to be careful!

Thanks I think I get it now. It made sense to give Q (heat energy) a negative sign since energy was dissipated. So the negative value indicates a loss in heat energy. Then working from there I got intial temp to be 428.81K

So it should've been (17+273.15) * (–99,000/1.4*510) = 428.81K

 

[jbriggs444]

Thanks I think I get it now. It made sense to give Q (heat energy) a negative sign since energy was dissipated. So the negative value indicates a loss in heat energy. Then working from there I got intial temp to be 428.81K

So it should've been (17+273.15) * (–99,000/1.4*510) = 428.81K

That formula is all messed up. It has multiplication instead of addition. And a sign error.

Note that -99,000/1.4*510 is also ambiguous. Are we supposed to multiply by the 510 or divide? The rules (PEMDAS or BEDMAS) prescribe multiplication. But it is pretty clear that we want division in this case.As I understand the problem we have a steel disk that dissipates 99 kJ of heat energy, ending at a temperature of 17 C. We are asked for the disk temperature prior to this loss of heat.

So we:
1. Convert the final temperature to Kelvin.
2. Compute the temperature difference due to a transfer of 99 kJ of heat energy.
3. Add the temperature difference to the final temperature to arrive at the initial temperature in Kelvin.
$$(17+273.15) + \frac{99000}{1.4 \times 510} = 428.81 \text{K}$$

[Two decimal places? Seriously? There is no justification for that excess precision. If we only know the final temperature to plus or minus 0.5 degrees C then we cannot reasonably judge the initial temperature to within plus or minus 0.005 K]