Question

a sheet of gold weighing 12.0 g and at a temperature of 13.6 °c is placed flat on a sheet of iron weighing 22.0 g and at a temperature of 53.4 °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 the 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 the surroundings. The formula for heat transfer is \(Q = mc\Delta T\), where \(m\) is mass, \(c\) is specific heat capacity, and \(\Delta T\) is the change in temperature. The specific heat capacity of gold (\(c_{gold}\)) is \(0.129\ J/g^\circ C\) and of iron (\(c_{iron}\)) is \(0.449\ J/g^\circ C\). Let the final temperature be \(T_f\). Then \(Q_{gold}=m_{gold}c_{gold}(T_f - T_{gold})\) and \(Q_{iron}=m_{iron}c_{iron}(T_{iron}-T_f)\). Setting them equal: \(m_{gold}c_{gold}(T_f - T_{gold})=m_{iron}c_{iron}(T_{iron}-T_f)\)

Step2: Plug in the values

\(m_{gold} = 12.0\ g\), \(T_{gold}=13.6^\circ C\), \(m_{iron}=22.0\ g\), \(T_{iron}=53.4^\circ C\), \(c_{gold}=0.129\ J/g^\circ C\), \(c_{iron}=0.449\ J/g^\circ C\)

$$12.0\times0.129\times(T_f - 13.6)=22.0\times0.449\times(53.4 - T_f)$$

Step3: Simplify the left and right sides

Left side: \(1.548\times(T_f - 13.6)=1.548T_f-21.0528\)
Right side: \(9.878\times(53.4 - T_f)=9.878\times53.4-9.878T_f = 527.5852-9.878T_f\)

Step4: Solve for \(T_f\)

$$1.548T_f-21.0528 = 527.5852-9.878T_f$$

Add \(9.878T_f\) to both sides: \(1.548T_f + 9.878T_f-21.0528=527.5852\)

$$11.426T_f-21.0528 = 527.5852$$

Add \(21.0528\) to both sides: \(11.426T_f=527.5852 + 21.0528=548.638\)

$$T_f=\frac{548.638}{11.426}\approx48.0^\circ C$$

(after considering significant digits)

Answer:

\(48.0\)