Two substances, gold and granite, initially at different temperatures, come into contact and reach thermal equilibrium. The mass of gold is 20.2 g and its initial temperature is 58.2 ∘C. The mass of granite is 29.1 g and its initial temperature is 20.1 ∘C. What is the final temperature of both substances at thermal equilibrium? (The specific heat capacity of gold is 0.128 J/g⋅∘C; the specific heat capacity of granite is 0.79 J/g⋅∘C.)
heat lost by one + heat gained by other = 0.
[mass Au x specific heat Au x (Tfinal-Tinitial)] + [mass granite x specific heat granite x (Tfinal-Tinitial)] = 0
Substitute and solve for only unknown, i.e., Tfinal.
To find the final temperature at thermal equilibrium, we can apply the principle of heat transfer.
The heat transferred by gold (Qgold) will be equal to the heat transferred by granite (Qgranite). The equation for heat transfer is given by:
Q = mcΔT
where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
For gold:
Qgold = mcΔT
Qgold = (20.2 g)(0.128 J/g∙∘C)(Tf - 58.2 ∘C) ------(1)
For granite:
Qgranite = mcΔT
Qgranite = (29.1 g)(0.79 J/g∙∘C)(Tf - 20.1 ∘C) ------(2)
Since Qgold = Qgranite, we can set equation (1) equal to equation (2):
(20.2 g)(0.128 J/g∙∘C)(Tf - 58.2 ∘C) = (29.1 g)(0.79 J/g∙∘C)(Tf - 20.1 ∘C)
Expanding and rearranging the equation gives:
0.128(Tf - 58.2) = 0.79(Tf - 20.1)
0.128Tf - 8.6784 = 0.79Tf - 15.879
0.79Tf - 0.128Tf = 15.879 - 8.6784
0.662Tf = 7.2006
Tf = 7.2006 / 0.662
Tf ≈ 10.9036 ∘C
Therefore, the final temperature of both substances at thermal equilibrium is approximately 10.9036 °C.
To find the final temperature at thermal equilibrium, we can apply the principle of heat transfer, which states that heat lost by one substance is equal to the heat gained by another substance.
The equation for heat transfer is given by:
q = mcΔT
Where:
q is the heat transferred
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
For gold, we can calculate the heat lost (q₁) using the formula:
q₁ = mcΔT
m₁ = 20.2 g (mass of gold)
c₁ = 0.128 J/g⋅∘C (specific heat capacity of gold)
ΔT₁ = T - 58.2 ∘C (change in temperature of gold)
For granite, we can calculate the heat gained (q₂) using the formula:
q₂ = mcΔT
m₂ = 29.1 g (mass of granite)
c₂ = 0.79 J/g⋅∘C (specific heat capacity of granite)
ΔT₂ = T - 20.1 ∘C (change in temperature of granite)
At thermal equilibrium, the heat lost by gold (q₁) is equal to the heat gained by granite (q₂). Therefore, we can set up an equation:
q₁ = q₂
m₁c₁ΔT₁ = m₂c₂ΔT₂
Now we can substitute the given values into the equation and solve for the final temperature (T).
20.2 g * 0.128 J/g⋅∘C * (T - 58.2 ∘C) = 29.1 g * 0.79 J/g⋅∘C * (T - 20.1 ∘C)
After simplifying and solving the equation, we find the final temperature (T) to be approximately 23.2 ∘C.