Question

A student places a 72.0 gram sample of metal at an initial temperature of 100.˚C in a coffee cup calorimeter that contains 40.0 grams of water at an initial temperature of 25.0 ˚C. After addition of the metal, the temperature of the water / metal mixture was monitored using a temperature probe. The highest temperature of the water / solid combination was found to be 35.0 ˚C. (The specific heat of water = 4.18 J / g ˚C )

Do not use spaces when giving your answers and remember significant figures

Useful equations: ΔT = Tfinal ̶ T initial ; q = specific heat × mass × ΔT

What was ΔT for the water? °C

What was ΔT for the metal? °C

How much heat ( q water ) did the water gain? J

Assuming no heat was lost to the surroundings, what is the value of q metal ? J

What is the specific heat of the metal? J / ( g °C )

Answer 1

Q = m c ∆T
Q = quantity of heat in joules (J)
m = mass of the substance acting as the environment in
grams (g)
c = specific heat capacity (4.19 for H2O) in J/(g ^oC)
∆T = change in temperature = Tfinal - Tinitial in ^oC

Heat was lost by metal = heat gained by water

Let us calculate heat gained by water

m = 40

c = 4.18

ΔT = 35-25 = 10

Q = 40 x 4.18 x 10 = 1672 Joules

1672 joules of heat was gained by water . .

specific heat of the metal

Q = 1672

m = 72 gm

C= ?

ΔT = 100 - 35 = 65

SUbstitute in the equation and find C

1672 = 72 x C x 65

C = 0.3573  J / ( g °C )

.The specific heat of the metal is 0.3573 J / ( g °C )