Question

a sheet of gold weighing 8.9 g and at a temperature of 17.1 °c is placed flat on a sheet of iron weighing 21.2 g and at a temperature of 50.7 °c. what is the final temperature of the combined metals? assume that no heat is lost to the surroundings. be sure your answer has the correct number of significant digits.

Explanation:

Step1: Recall Heat Transfer Formula

The heat gained by one substance equals the heat lost by the other (since no heat is lost to surroundings). The formula for heat transfer is \( q = mc\Delta T \), where \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is change in temperature. For gold (g) and iron (Fe), \( q_{gained} = -q_{lost} \), so \( m_gc_g(T_f - T_{g,i}) = -m_Fec_Fe(T_f - T_{Fe,i}) \). Specific heat capacities: \( c_g = 0.129 \, \text{J/g°C} \), \( c_{Fe} = 0.449 \, \text{J/g°C} \), \( m_g = 8.9 \, \text{g} \), \( T_{g,i} = 17.1 \, \text{°C} \), \( m_{Fe} = 21.2 \, \text{g} \), \( T_{Fe,i} = 50.7 \, \text{°C} \).

Step2: Substitute Values into Equation

\( 8.9 \times 0.129 \times (T_f - 17.1) = -21.2 \times 0.449 \times (T_f - 50.7) \)

Calculate left side coefficient: \( 8.9 \times 0.129 \approx 1.1481 \)

Right side coefficient: \( -21.2 \times 0.449 \approx -9.5188 \)

Equation becomes: \( 1.1481(T_f - 17.1) = -9.5188(T_f - 50.7) \)

Step3: Expand Both Sides

Left: \( 1.1481T_f - 1.1481 \times 17.1 \approx 1.1481T_f - 19.6325 \)

Right: \( -9.5188T_f + 9.5188 \times 50.7 \approx -9.5188T_f + 482.6032 \)

Step4: Solve for \( T_f \)

Bring all \( T_f \) terms to left and constants to right:

\( 1.1481T_f + 9.5188T_f = 482.6032 + 19.6325 \)

\( 10.6669T_f = 502.2357 \)

\( T_f = \frac{502.2357}{10.6669} \approx 47.1 \, \text{°C} \) (check significant digits: masses have 2/3 sig figs, temps have 3, specific heats have 3; result should have 3 sig figs)

Answer:

\( 47.1 \)