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

The correct answer is "5217 Cal".

Step-by-step 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* 3.76* (77-21.5)`

⇒
`=25* 3.76* 55.5`

⇒
`=5217 \ Cal`