Question

3. A 10 g sample was heated from 10°C to 81°C using 980J of energy, what is C of thematerial?

Answer 1

c = 1.380 J/g°C

Step-by-step explanation

Given:

Mass of sample, m = 10 g

Initial temperature, T₁ = 10 °C

Final temperature, T₂ = 81 °C

Change in temperature, ΔT = T₂ - T₁ = 81 °C - 10 °C = 71 °C

Quantity of heat, Q = 980J

What to find:

The specific heat, c of the material.

Step-by-step solution:

The specific heat, c of the material can be calculated using the formula given below:

`\begin{gathered} Q=mc\Delta T \\ \\ \Rightarrow c=(Q)/(m\Delta T) \end{gathered}`

Plugging the values of the parameter, we have;

`c=(980J)/(10g*71°C)=\frac{980J}{710\text{ }g°C}=1.380\text{ }J\text{/}g°C`

Therefore, the specific heat, c of the material is 1.380 J/g°C