A Styrofoam bucket of negligible mass contains 1.75 kg of water and 0.450 kg of ice. More ice, from a refrigerator at -15.0∘C, is added to the mixture in the bucket, and when thermal equilibrium has been reached, the total mass of ice in the bucket is 0.884 kg.
Assuming no heat exchange with the surroundings, what mass of ice was added?
Not sure I understand this. Yes energy will have to come from the water to raise the temp of the ice from -15 to 0, but it won't cause any phase change of the water , there's too much of it.
.884 - .450 = .434 kg. Unless I'm really missing something...
To determine the mass of ice that was added, we need to consider the principle of conservation of energy. Since there is no heat exchange with the surroundings, the heat lost by the colder objects (the added ice and the original ice) must equal the heat gained by the warmer objects (the water).
First, let's find the initial heat released by the ice that was already in the bucket. We can calculate this using the equation:
Q = m * c * ∆T
Where:
Q = Heat (in Joules)
m = Mass (in kg)
c = Specific heat capacity (for ice, it is approximately 2.09 J/g°C, or 2090 J/kg°C)
∆T = Change in temperature (in °C)
The initial temperature of the ice in the bucket is 0°C, and its final temperature is -15°C. To convert the specific heat capacity from grams to kilograms, we divide by 1000.
Q(initial ice) = m(initial ice) * c * ∆T(initial ice)
Q(initial ice) = 0.450 kg * 2090 J/kg°C * (0°C - (-15°C))
Now, let's find the heat gained by the water in the bucket:
Q(water) = m(water) * c(water) * ∆T(water)
The initial temperature of the water is also 0°C. The final temperature will be the same as the final temperature of both the ice and the added ice because thermal equilibrium has been reached.
Since the bucket is of negligible mass, the mass of water remains the same throughout the process:
m(water) = 1.75 kg
To find the specific heat capacity of water, we can use the value of 4.18 J/g°C, or 4180 J/kg°C.
Q(water) = 1.75 kg * 4180 J/kg°C * (∆T(final ice))
Now, assuming conservation of energy, the heat released by the initial ice and the heat gained by the water are equal:
Q(initial ice) = Q(water)
0.450 kg * 2090 J/kg°C * (0°C - (-15°C)) = 1.75 kg * 4180 J/kg°C * (∆T(final ice))
Simplifying the equation will give you the change in temperature, ∆T(final ice), of the additional ice:
33,712.5 J = 7315 J * (∆T(final ice))
Solving for ∆T(final ice), we find:
∆T(final ice) ≈ 4.61°C
Finally, to determine the mass of the added ice, we can use the equation:
m(added ice) = Q(initial ice) / (c * ∆T(final ice))
m(added ice) = 33,712.5 J / (2090 J/kg°C * 4.61°C)
Now, calculating the mass of the added ice gives us:
m(added ice) ≈ 3.11 kg
Therefore, the mass of ice that was added to the bucket is approximately 3.11 kg.