1 50.0 grams of an unknown metal is heated to 100.0°C and then placed in a calorimeter with 250.0 mL of water. If the final temperature of the
metal is 29.1°C and the calorimeter absorbed 824 Joules of energy, what
is the specific heat of the unknown metal? *

(a) Heat absorbed by the water

Δ

T

=

(

28.0

–

24.0

)

°

C

=

4.0

°

C

q

H₂O

=

m

c

Δ

T

=

100.0

g

×

4.184 J⋅

°C⁻¹g⁻¹

×

4.0

°C

=

1670 J

(b) Heat lost by the metal

Heat lost by metal = heat gained by water = 1670 J

(c) Heat capacity of the metal

q

metal

=

C

Δ

T

C

=

q

metal

Δ

T

=

1670 J

4.0 °C

=

420 J/°C

(d) Specific heat capacity of the metal

Specific heat capacity means the heat capacity per gram.

c

=

C

m

=

420 J⋅°C

−

1

50.0 g

=

8.4 J⋅°C

−

1

g

−

1

Explanation: