A 56.3g of water was heated to 51.2C and added to 45.8g of 20.3C water in a calorimeter.
Assume that no heat was lost to the surroundings; calculatee the final temp. of the resulting solution.

I really don't know how to do this..
...
I think I need 4.18J/g C , this number..
But this one is really confusing

Eliminating the verbiage that's not important, you have:

56.3 grams of water at 51.2 Celsius.

45.8 grams of water at 20.3 Celsius.

Let C be the heat capacity
Let T be the final temperature

Obviously, the 56.3 grams of water will have to cool and the 45.8 grams of water will have to warm up. The heat that is lost from the 56.3 grams of water will be added to the 45.8 grams of water.

Heat lost = 56.3 C(51.2-T)
Heat gained = 45.8 C(T-20.3)

Since heat lost by the hotter is equal to heat gained by the cooler water, you can equate those two expressions. When you do that, you'll find that the heat capacity can be eliminated from the equation. In other words, you don't really need to know that the heat capacity is 4.18J/g C -- as long as the heat capacities of the two liquids are the same.

You solve it from there. Basically, this boils down to a weighted average of the two temperatures. Take a mental note that the final temperature must be between 20.3C and 51.2C. If you get something outside of that range, you've made a mistake.

Typo Perito,


Heat lost = 56.3 C(51.2-T)
Heat gained = 45.8 C(T-45.8)

Should be "Heat gained = 45.8 C(T-20.3)"

Yup. I actually caught that when I was checking my answer -- then I forgot to change it in the post.

I got 37.3ºC

If the final temperature for above were 34.7ºC
a)how much heat was lost to the calorimeter?
b)calculate heat capacity
c)for the calorimeter.

so, a) is 37.3-34.7=2.6?

I don't know how to get a... please help me Thanks

Assume that no heat was lost to the surroundings


Because of the assumption, no heat was lost to the calorimeter. If you change that assumption (it sounds like the next part of the problem did that), you need to provide the mass of the calorimeter and its initial and final temperature. In a real calorimeter, you'd have to know its construction -- what insulation is present, etc.

With the information you've given, this is impossible to calculate.