Posted by **jan** on Friday, June 22, 2007 at 8:30pm.

If 500 kg of water at 200 degrees F is poured into a lake that is 50.0 degrees F, how much heat is added to the lake?

Joules added = mass x specific heat water x delta T.

mass = 500 kg

Look up specific heat water in cal/kg

delt T is 200.

Solve for Joules added.

Isn't the specific heat of water 1.00 cal/g?? If so, when I solved, I got 75000 k/cal. Is that correct?

The specific heat of water is 1.00 cal/g degC which equals 1.00 kcal/kg degC

I don't know how you obtained 7,500 cal and I don't know what unit k/cal is.

500 kg water = 500,000 grams.

1 cal/g*C is specific heat.

delta T = 200 F or 93.3

^{o}C.

okay,now I'm really confused. the k/cal was typed wrong, it should have been kcal or Kilocalorie. sorry :) however, you said the change in temp (delta T) was 200 F or 93.3 C which means of course that I need to convert from celsius to Fahrenheit, a step I missed.

But do I not need to take into consideration that the 500 kg of water that was 200 F is poured into the lake, the lake was 50 F. Do I not need to subtract the temperatures to get the change in temp or Delta T?

The way I read the problem, it asks only for the heat ADDED to the lake and I don't think the mass of water in the lake or the temperature of the lake has anything to do with how much heat was added from the 200 F water. Of course the temperature of the lake would change slightly (but only slightly) since it is so massive compared with the 500 kg of water added. Note that no mass of lake water (or volume) is given; therefore, it would be impossible to calculate a final T anyway. The cal = mass x specific heat x T is the heat contained in the 500 kg water at 200 F and that is what is added to the lake. Others may interpret the problem differently. I hope this helps.

how do you solve to get joules.

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