Homework Problem 8.21

[David Bui 1C]

A piece of copper of mass 20.0 g at 100.0 C is placed in a vessel of negligible heat capacity but containing 50.7 g of water at 22.0 C. Calculate the final temperature of the water. Assume
that no energy is lost to the surroundings.

I am assuming that this has something to do with the equation q=nCmΔT. I may be wrong.

[Cindy Chen_2I]

should be something like this:
heat released = heat absorbed
20 (mass)*0.39 (specific heat of Cu) *(100-T) (change in temperature) = 50.7 (mass)* 1 (specific heat of H2O) *(T-22)(change in temperature)

Try to solve for T. I got 32.4 degree Celsius as the final temperature.

[David Bui 1C]

should be something like this:
heat released = heat absorbed
20 (mass)*0.39 (specific heat of Cu) *(100-T) (change in temperature) = 50.7 (mass)* 1 (specific heat of H2O) *(T-22)(change in temperature)

Try to solve for T. I got 32.4 degree Celsius as the final temperature.

Thanks you gave me the right way to solve it. But the final answer was 25 degrees Celcius.

[Cindy Chen_2I]

Oh haha sorry. It should be 20*0.39*(100-T) = 50.7*4.18*(T-22)
you get 24.8 degree for the answer.

Sorry again for the confusion.

[KayleeMcCord1F]

I think the equation goes (Tf-Ti) so for the Copper side it should be (T-100), because both the water and Copper should have the same final temperature at equilibrium. This will yield the 24.8 degrees C.

[Edward Suarez 1I]

I think the equation goes (Tf-Ti) so for the Copper side it should be (T-100), because both the water and Copper should have the same final temperature at equilibrium. This will yield the 24.8 degrees C.

i thought the same but every time i worked it out i always got 19. something... am i doing the math part wrong lol

[Adrian C 1D]

No it would actually be (100-Tf). This is due to the temperature of the copper being 100 degrees celsius, which would result in the change in temperature for the copper to be negative. This would result in a negative answer when (Tf -100) or (100-Tf) to make it easier in the equation.

[JTieu_1L]

I think the equation goes (Tf-Ti) so for the Copper side it should be (T-100), because both the water and Copper should have the same final temperature at equilibrium. This will yield the 24.8 degrees C.

i thought the same but every time i worked it out i always got 19. something... am i doing the math part wrong lol

You were supposed to set them equal to each other not adding them up.

[AustinMcBrideDis3L]

I think I'm seeing why the previous explanations, thought mostly correct, aren't allowing you to arrive at the correct answer. Yes you set the equations equal to each other and it should always be TFinal-Tinitial so the unknown T should be first in both sides. However, to get the correct 25 (not 19) remember that q_released = -q_absorbed
This negative sign is crucial to getting the answer correct the right way.

[emmaferry2D]

I think I'm seeing why the previous explanations, thought mostly correct, aren't allowing you to arrive at the correct answer. Yes you set the equations equal to each other and it should always be TFinal-Tinitial so the unknown T should be first in both sides. However, to get the correct 25 (not 19) remember that q_released = -q_absorbed
This negative sign is crucial to getting the answer correct the right way.

ok that makes sense, I kept getting than answer 19 like the people above and was confused on what I was doing wrong. Thank you for the clarification, this was super helpful!

[sophiavmr]

The change in temperature is always Tfinal - Tinitial, so the values should be T-100 and T-22