A sheet of gold weighing 10.0 g and at a temperature of 18.0°C is placed on a sheet of iron weighing 20.0 g and at a temperature of 55.6°C. What is the final temperature of the combined metals? The specific heats of iron and gold are, respectively, 0.444 J/g°C and 0.129 J/g°C.

• A. 32.0°C
• B. 37.0°C
• C. 45.0°C
• D. 40.0°C

Answer: (B)

Verified step by step guidance

1. Identify the concept: This problem involves the principle of heat transfer between two substances until thermal equilibrium is reached. The heat lost by the hotter substance will be equal to the heat gained by the cooler substance.
2. Write the heat transfer equation: The heat gained or lost by a substance can be calculated using the formula: $q=mc\mathrm{\Delta T}$, where $q$ is the heat energy, $m$ is the mass, $c$ is the specific heat capacity, and $\mathrm{\Delta T}$ is the change in temperature.
3. Set up the equation for thermal equilibrium: Since the heat lost by iron will be equal to the heat gained by gold, we can set up the equation: $m_{\mathrm{Fe}}{c}_{\mathrm{Fe}}(T_{\mathrm{final}}-T_{\mathrm{initial,Fe}})\; =\; -m_{\mathrm{Au}}{c}_{\mathrm{Au}}(T_{\mathrm{final}}-T_{\mathrm{initial,Au}})$.
4. Substitute the known values: Plug in the masses, specific heat capacities, and initial temperatures for iron and gold into the equation. For iron: $m_{\mathrm{Fe}}=\; 20.0$ g, $c_{\mathrm{Fe}}=\; 0.444$ J/g°C, $T_{\mathrm{initial,Fe}}=\; 55.6$°C; for gold: $m_{\mathrm{Au}}=\; 10.0$ g, $c_{\mathrm{Au}}=\; 0.129$ J/g°C, $T_{\mathrm{initial,Au}}=\; 18.0$°C.
5. Solve for the final temperature: Rearrange the equation to solve for $T_{\mathrm{final}}$. This involves isolating $T_{\mathrm{final}}$ on one side of the equation and performing algebraic operations to find its value.