A piece of copper block of mass 24g at 230 degree celsius is placed in a copper calorimeter of mass 60g containing 54g of water at 31 degree celsius.Assuming heat losses are negligible,calculate the final steady temperature of the mixture

Question

(Specific heat capacity of water=4200J/kg/k)(Specific heat capacity /of copper=400J/kg/k)

Answer 1

.024 (400) (230 -T) = .060(400)(T-31) + .054(4200)(T-31)

Answer 2

to get the answer

Answer 3

Yes

Answer 4

Faisat

Answer 5

Heat lost= 0.024×400(230-T)
Heat gained by water= 0.05×4200(T-31)
Heat gained by calorimeter=0.060×400(T-31)
Heat lost= Heat gained
= 0.024×400(230-T) = 0.54×4200
(T-31)+0.60×400(T-31)
T= 38. 34°C