Question

A 50g chunk of metal (c=0.0506 j/g°C) is heated to 60°C. it is then dropped into a calorimeter(c=1.40 j/g°C,m=5.0 g)that contains 100.0g of an unknown liquid. the temperature of the liquid raises from 15°C to 20°C.What is the c of the calorimeter?

Answer 1

Isn't c given in the problem for the calorimeter as 1.40 J/g*C?

Answer 2

It's supposed to be "c" of liquid ,I'm sorry.

Answer 3

heat capacity of the calorimeter is 1.40 J/g*C x 5 g = 7.00 J/C

heat lost by metal + heat gained by calorimeter + heat gained by water = 0

[mass metal x specific heat metal x (Tfinal-Tinitial)] + [7.00 x (Tfinal-Tinitial)] + [mass liquid x specific heat liquid x (Tfinal-Tinitial)] = 0

note:Tf and Ti are the same for the calorimeter and liquid since the liquid is IN the calorimeter.

Substitute and solve for specific heat liquid.