Question

A sheet of gold weighing 11.4 g and at a temperature of 14.5°C is placed flat on a sheet of iron weighing 18.4 g and at a temperature of 55.4°C. What is the final temperature of the combined metals? Assume that no heat is lost to the surroundings.

Answer 1

The final temperature of the combined metals is 49.2314 °C

Step-by-step explanation:

Heat gain by gold = Heat lost by iron

Thus,

`m_(gold)* C_(gold)* (T_f-T_i)=-m_(iron)* C_(iron)* (T_f-T_i)`

Where, negative sign signifies heat loss

Or,

`m_(gold)* C_(gold)* (T_f-T_i)=m_(iron)* C_(iron)* (T_i-T_f)`

For gold:

Mass = 11.4 g

Initial temperature = 14.5 °C

Specific heat of gold = 0.129 J/g°C

For iron:

Mass = 18.4 kg

Initial temperature = 55.4 °C

Specific heat of water = 0.450 J/g°C

So,

`11.4* 0.129* (T_f-14.5)=18.4* 0.450* (55.4-T_f)`

`1.4706* (T_f-14.5)=8.28* (55.4-T_f)`

`1.4706* T_f-1.4706* 14.5=8.28* 55.4-8.28* T_f`

`1.4706* T_f-21.3237=458.712-8.28* T_f`

`1.4706* T_f+8.28* T_f=458.712+21.3237`

`T_f=49.2314`

Thus,

The final temperature of the combined metals is 49.2314 °C