Physics

Problem

200\,\text{g}

copper block at

150^{\circ}\text{C}

is dropped into

500\,\text{g}

of water at

25^{\circ}\text{C}

inside an insulated cup so that no energy enters or leaves the combined system. The specific heat capacities are

c_{\text{Cu}} = 0.39\,\text{J}\,\text{g}^{-1}\,^{\circ}\text{C}^{-1}

and

c_{\text{water}} = 4.18\,\text{J}\,\text{g}^{-1}\,^{\circ}\text{C}^{-1}

. Determine the approximate final equilibrium temperature of the system.

Intuitive explanation

In a thermally isolated system, energy is conserved—the heat lost by the hot copper must equal the heat gained by the cooler water. The key is to set up the heat balance equation $Q_{\text{lost}} = Q_{\text{gained}}$ using $Q = mc\Delta T$ , then solve for the final temperature where both substances reach the same value. Since copper has much lower heat capacity than water, it will cool significantly while the water warms only slightly.

Intuitive explanation

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