Question

The specific heat of a metal is 3.76 cal/g°C. How much heat energy will it take to heat a 25.0 g cylinder of the metal from 21.5°C to 77.0°C?

Answer 1

Answer:

The correct answer is "5217 Cal".

Explanation:

The given values are:

Specific heat,

c = 3.76 cal/g°C

Mass,

m = 25.0 g

Initial temperature,

T₁ = 21.5°C

Final temperature,

T₂ = 77.0°C

Now,

The heat energy will be:

⇒ $Q=mc \Delta t$

On substituting the given values, we get

⇒     $=25.0\times 3.76\times (77-21.5)$

⇒     $=25\times 3.76\times 55.5$

⇒     $=5217 \ Cal$