Question

the specific heat of a metal object is .21 cal/g C. The metal is heated to 96C then transferred to a calorimeter contain 75g of water at 18 C. The metal and water reach a final temperature of 22C. What is the mass of the metal?

a 4g

b 38 g

300 g

75 g

19 g

Answer 1

Heat Loss by metal = Heat Taken by water

+Q(water) = -Q(metal)

Q = mcΔT. ΔT = Tf - Ti

Initial temperature of metal, Ti = 96 ºC

Tf = 22 ºC

C = specific heat of metal = .21 cal/g C

Mass of metal be ‘a’ gram

Qm = a g x .21 cal/g C x (22 – 96) ºC

water, V = 75 mL = 75 g

Initial temperature of water, Ti = 18 ºC

Tf = 22 ºC

C = specific heat of copper = 1 cal/g°C

Qw = 75 g x 1 cal/g C x (18 – 22) ºC

+Q(water) = -Q(copper)

75 g x 1 cal/g C x (22 – 18) ºC = - [a g x .21 cal/g C x (22 – 96) ºC]

300 = 15.54 x a

a = 19 g