Question

Find the specific heat of a material if a 6-g sample absorbs 50 J when it is heated from 30°C to 50°C.

Answer 1

Answer:

The specific heat of the material is:

$\text{c = }416.67\text{ }Jkg^{-1\text{ 0}}C^{-1}$Explanations:

The mass of the sample, m = 6g = 6/1000 kg = 0.006 kg

m = 0.006 kg

The amount of heat absorbed, H = 50 J

Initial temperature, θ₁ = 30°C

Final temperature, θ₂ = 50°C

The Heat (H) absorbed by a material of mass (m) when heated from a temperature θ₁ to θ₂ is given as:

$\begin{gathered} H\text{ = mc(}\theta_{2}\text{ - }\theta_{1}) \\ \text{where c is the }specific\text{ heat capacity} \\ 50\text{ = 0.006 }\times\text{ c }\times\text{ (50 - 30)} \\ 50\text{ = 0.006 }\times\text{ c }\times\text{ 20} \\ 50\text{ = }0.12c \\ \text{c = }\frac{50}{0.12} \\ \text{c = }416.67\text{ }Jkg^{-1\text{ 0}}C^{-1}^{} \end{gathered}$