Question

You have 120 g of water in a 65 g calorimeter
cup at a temperature of 23◦C. You drop a
340 g piece of metal at a temperature of 120◦C
into the calorimeter cup and the temperature
stabilizes at 57◦C.
How much heat did the calorimeter cup
absorb? The specific heat of the water is
1 cal/g ·
◦ C and that of the calorimeter is
0.32 cal/g ·
◦ C.
Answer in units of cal.
008 (part 2 of 4) 3.0 points
How much heat did the water absorb?
Answer in units of cal.
009 (part 3 of 4) 3.0 points
How much heat did the metal lose?
Answer in units of cal.
010 (part 4 of 4) 3.0 points
What was the specific heat of the metal?
Answer in units of cal.

Answer 1

First part:
q = [mass cup x specific heat cup x (Tfinal-Tinitial)]
q = [65 x 0.32 cal/g x (57-23)]= ?

2nd part:
q = [mass water x specific heat H2O x (Tfinal-Tinitial)]
q = [120 g x 1 cal/g x (57 - 23)] = ?

3rd part:
The heat lost by the metal is the heat gained by the water + heat gained by the cup.

4th part:
q = mass metal x specific heat metal x (Tfinal-Tinitial)
q = 340 g x specific heat metal x (57 - 120) = ?
Solve for specific heat metal.
Post your work if you run into trouble with any part.