Question

How much thermal energy must be transferred to a 20.0-g piece of copper wire to raise it from room temperature (25.0°C) to its melting temperature (1085.0°

CCu = 385 J/(kg·°C)

Answer 1

We can find the quantity of the thermal energy from the formula:

$$Q = mc \Delta T,$$

here, $Q$ is the quantity of the thermal energy that must be transferred to a 20.0-g piece of copper wire to raise it from room temperature (25.0°C) to its melting temperature (1085.0°C), $m = 0.02 kg$ is the mass of the piece of copper wire, c = 385 \dfrac{J}{kg \cdot \! ^{\circ}C} is the specific heat capacity of the copper, $\Delta T$ is the change in temperature.

Then, we get:

Q = 0.02kg \cdot 385 \dfrac{J}{kg \cdot \! ^{\circ}C} \cdot (1085 \ ^{\circ}C - 25 \ ^{\circ}C) = 8162J.

Answer:

$Q = 8162J.$