Thermodynamics Challenge #4 — Specific Heat

The specific heat
of the pan.

Thermodynamics Easy

Problem

A 0.5 kg metal pan on the stove absorbs 9,000 J of energy, and its temperature rises by 40 °C. Calculate the specific heat capacity of the metal.

Pan mass\( m = 0.5\ \mathrm{kg} \)
Energy absorbed\( Q = 9000\ \mathrm{J} \)
Temperature rise\( \Delta T = 40\ ^{\circ}\mathrm{C} \)
Answer inJ/(kg·°C)

All the energy goes into the pan — no losses.

Hint: same equation, different unknown: \( Q = mc\Delta T \) rearranged for \( c \). Divide the energy by mass times temperature rise.

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The specific heatof the pan.

Problem