a 500g piece of silver at 150 degrees celsuis is submerged in 1000g of water at 5 degrees celsuis to be cooled. determine the final temperature of the silver and water given. Cwater=4.18 x 10^3 J/kg.C Csilver = 2.4 x 10^2 J/kg.C
Set the heat loss of the silver equal to the heat gain by the water. The final temperatgure should be the only unknown in the equation. Solve for it.
See if you can set up the equation yourself. Heat loss when going from T1 to T2 is M C (T1 - T2)
but do i do sperate equations for both water and silver or do i do it as a combined equation. if its combined do i combine the masses and he heat capacity? i'm so confused
To determine the final temperature of the silver and water after they reach thermal equilibrium, we can use the principle of conservation of energy.
The heat lost by the silver (Qsilver) will be equal to the heat gained by the water (Qwater). We can calculate the heat lost/gained using the formula:
Q = mcΔT
Where:
- Q is the heat transferred (in joules)
- m is the mass of the substance (in kg)
- c is the specific heat capacity of the substance (in J/kg°C)
- ΔT is the change in temperature (in °C)
Let's calculate the heat lost by the silver and the heat gained by the water separately:
For the silver:
m = 500g = 0.5 kg (mass of silver)
c = 2.4 x 10^2 J/kg°C (specific heat capacity of silver)
ΔT = Tf - Ti (final temperature of the silver - initial temperature of the silver)
ΔT = Tf - 150°C
Qsilver = mcΔT
Qsilver = 0.5 kg * (2.4 x 10^2 J/kg°C) * (Tf - 150°C)
For the water:
m = 1000g = 1 kg (mass of water)
c = 4.18 x 10^3 J/kg°C (specific heat capacity of water)
ΔT = Tf - Ti (final temperature of the water - initial temperature of the water)
ΔT = Tf - 5°C
Qwater = mcΔT
Qwater = 1 kg * (4.18 x 10^3 J/kg°C) * (Tf - 5°C)
Since Qsilver = Qwater (heat lost = heat gained), we can equate the two equations:
0.5 kg * (2.4 x 10^2 J/kg°C) * (Tf - 150°C) = 1 kg * (4.18 x 10^3 J/kg°C) * (Tf - 5°C)
Now, we can solve this equation for Tf. Let's calculate it:
0.5 * (2.4 x 10^2) * (Tf - 150) = 1 * (4.18 x 10^3) * (Tf - 5)
120 * (Tf - 150) = 4.18 * (Tf - 5)
120Tf - 120 * 150 = 4.18Tf - 4.18 * 5
120Tf - 120 * 150 = 4.18Tf - 20.9
120Tf - 18000 = 4.18Tf - 20.9
(120Tf - 4.18Tf) = (18000 - 20.9)
115.82Tf = 17979.1
Tf = 17979.1 / 115.82
Tf ≈ 155.26°C
Therefore, the final temperature of the silver and water when they reach thermal equilibrium is approximately 155.26°C.