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# Specific Heat Capacity Question Watch

1. Hi! I'm having a bit of a problem with my Physics homework. It is day two, I have had two people explain it to me, and I am still very confused.
----- I am using an asterisk in place of the degree sign to make it clearer.-----

The question:

"In an experiment, 20.0g of hot sea water at 65.0*C is mixed with 80.0g of tap water at 12.0*C inside a copper calorimeter of mass 75.0g also at 12.0*C. If the thermal energy lost to the surroundings in negligible, calculate the new temperature of the mixture and the calorimeter. The specific heat capacities of sea water, tap water and copper are 3990 J/kg K, 4200 J/kg K, and 386 J/kg K respectively."

What I've done so far:

First I converted all temperatures to Kelvin, and all masses to kilograms.
The equation I need to use is Q = mc(final temp - initial temp).

Sea water:
Q = 79.8(final temp) - 26972.4 (after expanding the brackets)
final temp = (Q + 26972.4) / 79.8

Tap water:
Q = 336(final temp) - 95760
final temp = (Q + 95760) / 336

Copper:
Q = 28.95(final temp) - 8250.75
final temp = (Q + 8250.75) / 28.95

I have no idea where to go from there. The final temperature is obviously going to be the same for all of them, but I don't know what to do about Q. The friend who explained it to me said that Q "doesn't matter", but it's in the way and I can't do anything with the equations. Please help!

Thank you
2. (Original post by JustJusty)
Hi! I'm having a bit of a problem with my Physics homework. It is day two, I have had two people explain it to me, and I am still very confused.
----- I am using an asterisk in place of the degree sign to make it clearer.-----

The question:

"In an experiment, 20.0g of hot sea water at 65.0*C is mixed with 80.0g of tap water at 12.0*C inside a copper calorimeter of mass 75.0g also at 12.0*C. If the thermal energy lost to the surroundings in negligible, calculate the new temperature of the mixture and the calorimeter. The specific heat capacities of sea water, tap water and copper are 3990 J/kg K, 4200 J/kg K, and 386 J/kg K respectively."

What I've done so far:

First I converted all temperatures to Kelvin, and all masses to kilograms.
The equation I need to use is Q = mc(final temp - initial temp).

Sea water:
Q = 79.8(final temp) - 26972.4 (after expanding the brackets)
final temp = (Q + 26972.4) / 79.8

Tap water:
Q = 336(final temp) - 95760
final temp = (Q + 95760) / 336

Copper:
Q = 28.95(final temp) - 8250.75
final temp = (Q + 8250.75) / 28.95

I have no idea where to go from there. The final temperature is obviously going to be the same for all of them, but I don't know what to do about Q. The friend who explained it to me said that Q "doesn't matter", but it's in the way and I can't do anything with the equations. Please help!

Thank you
The values of Q are different.

Make use of conservation of energy.

Thermal energy gained by the tap water and copper calorimeter is equal to the thermal energy lost by the hot sea water.

In this equation, there is only one unknown which is the final temperature.
3. (Original post by Eimmanuel)
The values of Q are different.

Make use of conservation of energy.

Thermal energy gained by the tap water and copper calorimeter is equal to the thermal energy lost by the hot sea water.

In this equation, there is only one unknown which is the final temperature.
Thank you!

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