Question

A 23.9 g sample of an unknown metal is heated from 41.2C to 68.9C. During the process the

metal absorbs 1110 Joules. Calculate the specific heat of the metal.

Answer 1

1.677 J/gºC

Step-by-step explanation:

From the question given above, the following data were obtained:

Mass (M) = 23.9 g

Initial temperature (T₁) = 41.2 °C

Final temperature (T₂) = 68.9 °C

Heat (Q) absorbed = 1110 J

Specific heat capacity (C) of metal =?

Next, we shall determine the change in the temperature of the metal. This can be obtained as follow:

Initial temperature (T₁) = 41.2 °C

Final temperature (T₂) = 68.9 °C

Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 68.9 – 41.2

ΔT = 27.7 °C

Finally, we shall determine the specific heat capacity of the metal as follow:

Mass (M) = 23.9 g

Heat (Q) absorbed = 1110 J

Change in temperature (ΔT) = 27.7 °C

Specific heat capacity (C) of metal =?

Q = MCΔT

1110 = 23.9 × C × 27.7

1110 = 662.03 × C

Divide both side by 662.03

C = 1110 / 662.03

C = 1.677 J/gºC

Therefore, the specific heat capacity of the metal is 1.677 J/gºC