0.2 kg of ice at 0 degrees celcius is mixed with 500g of water at 25 degrees celcius. What is the resulting temperature?
To find the resulting temperature after mixing ice and water, we can use the principle of conservation of energy. The heat gained by the water should be equal to the heat lost by the ice.
First, let's calculate the heat gained by the water using the formula:
Q = mcΔT
where:
Q is the heat gained by the water,
m is the mass of the water (500g),
c is the specific heat capacity of water, which is approximately 4.18 J/g°C, and
ΔT is the change in temperature of the water (resulting temperature - initial temperature).
Q = (500g) * (4.18 J/g°C) * (T - 25°C)
Next, let's calculate the heat lost by the ice using the formula:
Q = mlf
where:
Q is the heat lost by the ice,
m is the mass of the ice (0.2kg),
lf is the latent heat of fusion of ice, which is approximately 334 J/g.
Q = (0.2kg) * (1000g/kg) * (334 J/g)
Since the heat gained and the heat lost are equal, we have the equation:
(500g) * (4.18 J/g°C) * (T - 25°C) = (0.2kg) * (1000g/kg) * (334 J/g)
Now, we can calculate the resulting temperature (T) by solving the equation for T:
(500g) * (4.18 J/g°C) * (T - 25°C) = (0.2kg) * (1000g/kg) * (334 J/g)
Simplify the equation:
(500g) * (4.18 J/g°C) * T - (500g) * (4.18 J/g°C) * 25°C = (0.2kg) * (1000g/kg) * (334 J/g)
(500g) * (4.18 J/g°C) * T - 500g * 104.5 J/°C = 66800 J
(500g) * (4.18 J/g°C) * T = 66800 J + 500g * 104.5 J/°C
Multiply and divide to isolate T:
(500g) * (4.18 J/g°C) * T = 66800 J + 500g * 104.5 J/°C
T = (66800 J + 500g * 104.5 J/°C) / (500g * 4.18 J/g°C)
Now, let's plug in the values and calculate the resulting temperature:
T = (66800 J + 500g * 104.5 J/°C) / (500g * 4.18 J/g°C)
T ≈ 7.86°C
Therefore, the resulting temperature after mixing ice at 0°C with water at 25°C is approximately 7.86°C.