Question

A sheet of gold weighing 10.3 g and at a temperature of 10.0°C is placed flat on a sheet of iron weighing 18.7 g and at a temperature of 57.9°C. What is the final temperature of the combined metals? Assume that no heat is lost to the surroundings.

Answer 1

Heat gained by sheet of gold=Heat lost by sheet of iron

Heat gained by gold = m_gold*c_gold*deltaT_gold

m_gold =mass of gold sheet = 10.3g

c_gold = heat capacity of gold = 0.129J/g°C

DeltaT_gold = (T_f - 10℃) ( let Tf be the final temperature)

Similarly, heat lost by iron = m_iron*c_iron*deltaT_iron

m_iron = mass of iron sheet= 18.7g

c_iron = heat capacity if iron = 0.450J/g℃

DeltaT_iron = (57.9 - Tf)℃

So,.

10.3g*0.129J/g℃*(Tf-10)℃ = 18.7g*0.450J/g℃*(57.9-Tf)℃

1.329J/℃(Tf-10)℃ = 8.415J/℃(57.9-Tf)℃

1.329Tf - 13.29℃ = 487.23℃ - 8.415Tf

Tf = 500.5/9.744℃

Tf = 51.4℃ ( Final temperature)