A piece of copper at 280 C is dropped into 1.1 kg of water at 25 C.
If the equilibrium temperature is 30 C, what is the mass of the copper?
sum of heats gained=0
masscopper*Ccu*(30-25)+1.1*Cwater*(30-280)=0
solve for masscopper
To solve this problem, we can use the principle of heat transfer, which states that the heat gained by one object equals the heat lost by another object.
First, let's calculate the heat lost by the copper. We can use the specific heat formula, which is given by:
Q = mcΔT
Q = heat lost by the copper
m = mass of the copper (unknown)
c = specific heat capacity of copper (0.39 J/g°C)
ΔT = change in temperature of the copper (280°C - 30°C = 250°C).
Q = mcΔT
Q = m * 0.39 * 250
Next, let's calculate the heat gained by the water. We can use the formula:
Q = mcΔT
Q = heat gained by the water
m = mass of the water (1.1 kg)
c = specific heat capacity of water (4.18 J/g°C)
ΔT = change in temperature of the water (30°C - 25°C = 5°C).
Q = mcΔT
Q = 1.1 * 4.18 * 5
Since we're dealing with an isolated system, the heat lost by the copper equals the heat gained by the water:
m * 0.39 * 250 = 1.1 * 4.18 * 5
Now we can solve for m, the mass of the copper:
m = (1.1 * 4.18 * 5) / (0.39 * 250)
m ≈ 0.114 kg
Therefore, the mass of the copper is approximately 0.114 kg.