Respuesta :
Answer:
[tex]T = 36.393\,^{\textdegree}C[/tex]
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:
[tex]-Q_{iron} = Q_{gold}[/tex]
[tex]-(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)[/tex]
[tex]-(3.616\,\frac{J}{^{\textdegree}C})\cdot (T-54.4\,^{\textdegree}C) = (2.515\,\frac{J}{^{\textdegree}C})\cdot (T-10.5^{\textdegree}C)[/tex]
[tex]-1.438\cdot (T - 54.4^{\textdegree}C) = T-10.5^{\textdegree}C[/tex]
[tex]-1.438\cdot T +78.227^{\textdegree}C = T - 10.5^{\textdegree}C[/tex]
[tex]2.438\cdot T = 88.727\,^{\textdegree}C[/tex]
The final temperature is:
[tex]T = 36.393\,^{\textdegree}C[/tex]
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
