Question

A sheet of gold weighing 8.8 g and at a temperature of 10.5°C is placed flat on a sheet of iron weighing 19.5 g and at a temperature of 54.4°C. What is the final temperature of the combined metals? Assume that no heat is lost to the surroundings.

Answer 1

Answer:

$T = 36.393\,^{\textdegree}C$

Explanation:

The contact between the sheet of gold and the sheet of iron allows a heat transfer until thermal equilibrium is done, which means that both sheets have the same temperature:

$-Q_{iron} = Q_{gold}$

$-(0.008\,kg)\cdot (452\,\frac{J}{kg\cdot ^{\textdegree}C} )\cdot (T-54.4\,^{\textdegree}C) = (0.0195\,kg)\cdot (129\,\frac{J}{kg\cdot ^{\textdegree}C} )\cdot (T-10.5\,^{\textdegree}C)$

$-(3.616\,\frac{J}{^{\textdegree}C})\cdot (T-54.4\,^{\textdegree}C) = (2.515\,\frac{J}{^{\textdegree}C})\cdot (T-10.5^{\textdegree}C)$

$-1.438\cdot (T - 54.4^{\textdegree}C) = T-10.5^{\textdegree}C$

$-1.438\cdot T +78.227^{\textdegree}C = T - 10.5^{\textdegree}C$

$2.438\cdot T = 88.727\,^{\textdegree}C$

The final temperature is:

$T = 36.393\,^{\textdegree}C$

Answer 2

Answer:

Final temperature of the combined metals = 49.3 °C

Explanation:

Data

gold mass, Mg = 8.8 g

initial gold temperature, Tg1 = 10.5 °C

iron mass, Mi = 19.5 g

initial iron temperature, Ti1 = 54.4 °C

final temperature for both materials, T2 = ? °C

gold specific heat, Cg = 0.129 J/(g °C)

iron specific heat, Ci = 0.444 J/(g °C)

The heat is transferred from the iron sheet, which is initially at a higher temperature, to the gold sheet, which is initially at a lower temperature.

Heat transferred to the gold sheet:

Q = Mg*Cg*(T2 - Tg1)

Heat transferred from the iron sheet:

Q = Mi*Ci*(Ti1 - T2)

Then:

Mg*Cg*(T2 - Tg1) = Mi*Ci*(Ti1 - T2)

Mg*Cg*T2 - Mg*Cg*Tg1 = Mi*Ci*Ti1 - Mi*Ci*T2

T2*(Mg*Cg + Mi*Ci) = Mi*Ci*Ti1 + Mg*Cg*Tg1

T2 = (Mi*Ci*Ti1 + Mg*Cg*Tg1)/(Mg*Cg + Mi*Ci)

T2 = (19.5*0.444*54.4 + 8.8*0.129*10.5)/(8.8*0.129 + 19.5*0.444)

T2 = 49.3 °C