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

Answer:

$C=0.662\frac{J}{g\°C}$

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:

$C=\frac{Q}{m\Delta T}$

Thus, by plugging in we obtain:

$C=\frac{-68.5J}{4.5g*(25-48)\°C} \\\\C=0.662\frac{J}{g\°C}$

Best regards!