Question

A piece of unknown metal with mass 68.6 g is heated to an initial temperature of 100 °C and dropped into 84 g of water (with an initial temperature of 20 °C) in a calorimeter. The final temperature of the system is 52.1°C. The specific heat of water is 4.184 J/g*⁰C. What is the specific heat of the metal?
A) 0.171
B) 0.343
C) 1.717
D) 3.433

Answer 1

The heat absorbed by water will be equal to heat given by the metal

The heat absorbed by water = Q = msΔt

Where

m = mass of water

s = specific heat of water = 4.18 J g⁻¹°C⁻¹

Δt = final temperature - initial temperature = (52.1-20)°C = 32.1°C

So heat absorbed by water = 84 X 4.184 X 32.1 = 11281.74 J

Heat given by metal

    = mass of metal X specific heat of metal X change in temperature

heat given by metal = 68.6 X specific heat of metal X (100-52.1)

11281.74 J = 68.6 X specific heat of metal X (100-52.1)

specific heat of metal = 3.43 J g⁻¹°C⁻¹

Answer 2

The piece of unknown metal is in thermal equilibrium with water such that Q of metal is equal to Q of the water. We write this equality as follows:

-Qm = Qw
Mass of metal (Cm)(ΔT) = Mass of water (Cw) (ΔT)

where C is the specific heat capacities of the materials.

We calculate as follows:

-(Mass of metal (Cm)(ΔT)) = Mass of water (Cw) (ΔT)
-68.6 (Cm)(52.1 - 100) = 42 (4.184) (52.1 - 20)
Cm = 1.717 -----> OPTION C