A 10. g cube of copper at a temperature \(T_1\) is placed in an insulated cup containing 10. g of water at a temperature \(T_2\). If \(T_1 \gt T_2\), which of the following is true of the system when it has attained thermal equilibrium? (The specific heat of copper is \(\pu{0.385 J/(g}\cdot^\circ \)C) and the specific heat of water is \(\pu{4.18 J/(g}\cdot^\circ \)C).)

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The question mentions that the initial temperature of copper is greater (\(T_1 \gt T_2\)). This implies that the copper will transfer heat to the water. Since copper has a lower specific heat, less energy is required to change the temperature of copper. Because both the copper and water have equal masses, the temperature of copper must have decreased much more than the increase in temperature of the water.

$$ q_{\text{lost by copper}} = -q_{\text{gained by water}} $$ $$ mc\Delta T = mc\Delta T $$ $$ (10)(0.385)(T_1-T_f) = -(10)(4.18)(T_f-T_2) $$ $$ (0.385)(\Delta T_{\text{copper}}) = (4.18)(\Delta T_{\text{water}}) $$ $$ \Delta T_{\text{copper}} \gg \Delta T_{\text{water}} $$