Question

You heat up 225g of an unknown metal and place it in the water of a calorimeter. As the metal cools by 25.0 degrees C it raised the temperature of the water 12.5 C and then remains constant. What is the specific heat of the metal if the metal was place in 125g of water

Answer 1

the formula to use--->  specific heat= heat/ (mass x ΔT)

mass= 225 grams
ΔT= 25.0 C
heat= ? (missing)

in calorimeter problems, the concept that we have to understand is that heat (energy) is not formed or created but transferred. In other words, the heat lost by the unknown metal is the same amount of heat gained by the water. So, if we calculate the heat gain by the water is the same heat of the metal and we can plug it into our formula. 

heat= mass x specific heat x ΔT

mass= 125 grams
specific heat= 4.184 joules/gC or 1 calories/gC
ΔT= 12.5C

heat (in joules)= (125 grams)x(4.184 j/gC)x(12.5C)= 6540 joules
heat (in calories) (125 grams)X(1 cal/gC)x(12.5C)= 1560 calories

now we can use these values for the heat of the metal in the specific heat formula:

specific heat of metal= 6540 joules/ (225 grams x 25.0C)= 1.16 j/gC

or in calories

specific heat of metal= 1560 calories/ (225 grams x 25.0C)= 0.277 cal/gC