Question

A4.5 g metal sample decreases in temperature from 48°C to 25°C when 68.5

of heat is removed from it. Calculate the specific heat capacity of the metal.

Answer 1

`C=0.662(J)/(g\°C)`

Step-by-step explanation:

Hello!

In this case, since the heat removal when cooling down a body is computed via:

`Q=mC\Delta T`

Given the removed heat, mass and temperature change, we solve for the heat capacity of the metal as shown below:

`C=(Q)/(m\Delta T)`

Thus, by plugging in we obtain:

`C=(-68.5J)/(4.5g*(25-48)\°C) \\\\C=0.662(J)/(g\°C)`

Best regards!