Question

a sheet of gold weighing 9.7 g and at a temperature of 19.2 °c is placed flat on a sheet of iron weighing 21.1 g and at a temperature of 56.7 °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 gold ($Q_{gold}$) is equal to the heat lost by iron ($Q_{iron}$) 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. Specific heat capacities: $c_{gold} = 0.129\ J/g^\circ C$, $c_{iron} = 0.449\ J/g^\circ C$. Let final temperature be $T_f$.
For gold: $Q_{gold}=m_{gold}c_{gold}(T_f - T_{gold})$
For iron: $Q_{iron}=m_{iron}c_{iron}(T_{iron}-T_f)$
Set $Q_{gold}=Q_{iron}$: $m_{gold}c_{gold}(T_f - T_{gold}) = m_{iron}c_{iron}(T_{iron}-T_f)$

Step2: Plug in values

$m_{gold}=9.7\ g$, $T_{gold}=19.2^\circ C$, $m_{iron}=21.1\ g$, $T_{iron}=56.7^\circ C$, $c_{gold}=0.129$, $c_{iron}=0.449$
$9.7\times0.129\times(T_f - 19.2)=21.1\times0.449\times(56.7 - T_f)$

Step3: Simplify left and right sides

Left: $9.7\times0.129 = 1.2513$; so $1.2513(T_f - 19.2)=1.2513T_f - 1.2513\times19.2\approx1.2513T_f - 24.025$
Right: $21.1\times0.449\approx9.4739$; so $9.4739(56.7 - T_f)=9.4739\times56.7 - 9.4739T_f\approx537.17 - 9.4739T_f$

Step4: Solve for $T_f$

Set left = right: $1.2513T_f - 24.025 = 537.17 - 9.4739T_f$
Add $9.4739T_f$ to both sides: $1.2513T_f + 9.4739T_f - 24.025 = 537.17$
$10.7252T_f - 24.025 = 537.17$
Add 24.025 to both sides: $10.7252T_f = 537.17 + 24.025 = 561.195$
Divide by 10.7252: $T_f=\frac{561.195}{10.7252}\approx52.3^\circ C$

Answer:

$\boxed{52.3}$