Question

You are given a metal sample of weight 28.4 and asked to determine its specific heat. You add 1.25 x 104 J of heat to the sample and then measure a temperature increase of 18.0°C. What is the specific heat of the sample?

Answer 1

Answer:

$24.45 J/(kg^{\circ}C)$

Explanation:

The heat added to the sample is related to its increase in temperature by:

$Q=m C_s \Delta T$

where

$Q=1.25 \cdot 10^4 J$ is the heat added

$m=28.4 kg$ is the mass of the sample

$C_s$ is the specific heat

$\Delta T=18^{\circ}C$ is the increase in temperature

Substituting the numbers into the equation, we find the specific heat:

$C_s=\frac{Q}{m \Delta T}=\frac{1.25\cdot 10^4 J}{(28.4 kg)(18.0^{\circ}C)}=24.45 J/(kg^{\circ}C)$

Answer 2

Answer:

The specific heat of the metal is 24.5 J/g.°C.

Explanation:

Given data

• Mass of the metal (m): 28.4 g

• Heat added (Q): 1.25 × 10⁴ J

• Change in the temperature (ΔT): 18.0°C

• Specific heat of the metal (c): to be determined

We can determine the specific heat of the metal using the following expression.

Q = c × m × ΔT

c = Q / m × ΔT

c = 1.25 × 10⁴ J / 28.4 g × 18.0°C = 24.5 J/g.°C

The specific heat of the metal is 24.5 J/g.°C.