Question

A 200. g piece of metal is boiled in ethanol for 3 minutes and then transferred to a styrofoam cup containing 250. mL of water at 23 C. The metal and water are allowed to come to thermal equilibrium at a temperature of 27.5 C. If ethanol boils at a temperature of 78 C:

a. What is the delta T for the metal

b. What is the delta T for the water

c. what is the specific heat capacity for the unknown metal?

Answer 1

ΔT for the metal = 50.5 °C

ΔT for the water = 4.5 °C

The specific heat capacity for the unknown metal : 0.466 J/g° C

Further explanation

The law of conservation of energy can be applied to heat changes, i.e. the heat received/absorbed is the same as the heat released

Q in = Q out

Heat can be calculated using the formula:

Q = mc∆T

Q = heat, J

m = mass, g

c = specific heat, joules / g ° C

∆T = temperature difference, ° C / K

A 200. g piece of metal is boiled in ethanol (boiling point: 78 C), so :

m metal = 200 g

Ti metal(initial temperature of metal)=78 C

T(system temperature at equilibrium)=27.5

m water = 250 ml x 1 g/ml = 250 g

c water = 4.18 joules / g ° C

Tiw(initial temperature of water) = 23

• a. ΔT metal

`\tt 78-27.5=50.5^oC`

• b. ΔT water

`\tt 27.5-23=4.5^oC`

• c. the specific heat capacity for the unknown metal

`\tt Q~metal=Q~water\\\\200* c* (78-27.5)=250* 4.18* (27.5-23)\\\\10100* c=4702.5\\\\c=0.466~J/g^oC`