J.R. S. answered • 10/20/20
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The heat LOST by the metal must be equal to the heat GAINED by the water
heat = q = mC∆T where m=mass; C=specific heat; ∆T=change in temperature
heat lost by metal = (28 g)(C)(100 - 27) = (28g)(C)(73º) = 2044C
heat gained by water = 75 J/º x 7 = 525J
heat lost = heat gained so 2044C = 525
C = specific heat of the metal = 0.26 J/g/º
J.R. S.
tutor
Absolutely correct. Thanks for the catch. Made the correction.
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12/13/22
Gigi B.
Don't you need to know the mass of the water in order to do the second step - the heat gained by water?
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09/28/23
J.R. S.
tutor
@Gigi B.: Yes, normally that would be the case, but since they didn't provide that information, I assumed that the mass of water was included in the calorimeter constant of 75 J/º
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09/28/23
Colin S.
This is incorrect, J.R. The specific heat of the metal is 0.26 J/g*C. The heat gained by the water was 7 degrees centigrade, T''C-T'C, which is 27C-20C=7C. Units and significance withheld: 28*73=2044 75*7=525 525/2044~=0.2568512/13/22