what is the final temperature when a 3.0 kg gold bar at 99 degrees is dropped into a .22 kg of water at 25 degrees
47c
no
To determine the final temperature when a gold bar is dropped into water, we can use the principle of heat transfer, which states that the heat lost by the gold bar is equal to the heat gained by the water.
Let's start by calculating the amount of heat lost by the gold bar:
Q(bar) = m(bar) * c(bar) * ΔT(bar)
Where:
Q(bar) is the heat lost by the gold bar
m(bar) is the mass of the gold bar (3.0 kg)
c(bar) is the specific heat capacity of gold (0.129 J/g°C or 129 J/kg°C, since 1 kg = 1000 g)
ΔT(bar) is the change in temperature of the gold bar (final temperature - initial temperature)
Next, we can calculate the amount of heat gained by the water:
Q(water) = m(water) * c(water) * ΔT(water)
Where:
Q(water) is the heat gained by the water
m(water) is the mass of the water (0.22 kg)
c(water) is the specific heat capacity of water (4.18 J/g°C or 4180 J/kg°C, since 1 kg = 1000 g)
ΔT(water) is the change in temperature of the water (final temperature - initial temperature)
Since the two objects exchange heat until they reach thermal equilibrium, the amount of heat lost by the gold bar is equal to the amount of heat gained by the water:
Q(bar) = Q(water)
m(bar) * c(bar) * ΔT(bar) = m(water) * c(water) * ΔT(water)
Now, we can rearrange the formula to solve for the final temperature (ΔT(bar)):
ΔT(bar) = (m(water) * c(water) * ΔT(water)) / (m(bar) * c(bar))
Substituting the given values into the equation:
ΔT(bar) = (0.22 kg * 4180 J/kg°C * (final temperature - 25°C)) / (3.0 kg * 129 J/kg°C)
Simplifying the equation:
ΔT(bar) = (918.8 * (final temperature - 25)) / 387
To find the final temperature, we need to solve for ΔT(bar) when Q(bar) = Q(water), which means ΔT(bar) = 0. We can set the equation equal to zero:
0 = (918.8 * (final temperature - 25)) / 387
Simplifying the equation:
0 = 918.8 * (final temperature - 25)
Dividing both sides by 918.8:
0 = final temperature - 25
Adding 25 to both sides:
25 = final temperature
Therefore, the final temperature of the gold bar and water mixture is 25°C.
Heat will flow from the gold to the water until an equilbrium temperature T is reached. You will need the specific heats of gold (C1) and water (C2 = 1.00 kCal/kg C).
M2*C2*(99 - T) = M1*C1*(T - 25)
(99-T)/(T-25) = M1*C1/(M2*C2)
Solve for T (the final temperature). m1 is the mass of gold and M2 is the mass of water.