Question

A jeweler heats 20 g of gold until it has just melted. The temperature of the liquid gold is equal to the melting point (1336 K) and it is now poured over 40 g of silver contained in a container at the temperature of 293 K. Assume that heat is not exchanged with ambient or container and determine the common final temperature and state of respectively. silver and gold when thermal equilibrium has occurred.

Answer 1

When two substances at different temperatures are mixed together, then exchange of heat take place till temperature of both the substance become equal. This temperature is called as final Temperature of the mixture or equilibrium temperature.

If no heat loss to surrounding or with the container then by conservation of energy we can say that -

Heat given by one substance = Heat taken by another substance. - eq 1

Heat (H) = m*s*∆T

m = mass in kg

s= specific heat ( J / kg. K)

∆T=change in temperature

Specific heat of a substance is equal to heat gained or released by that substance to raise or fall the temperature of the substance by 1°C for a unit mass of a substance.

eq 1

m1*s1*(T1 - T) = m2*s2*(T - T2)

m1= mass of gold = 20 g =0.020kg

m2 = mass of silver= 40 g = 0.040kg

s1= specific heat of gold= 130 J/ kg K

s2 = specific heat of silver=236 J/kg K

As specific heat of gold and silver is not given in the question so i checked on internet .

T1= temperature of gold= 1336K

T2 =temperature of silver=293K

T =final temperature of the mixture= ?

0.020*130*(1336 - T) = 0.040*236*(T - 293)

After calculation, T = 518.232K