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.

Question

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?

Answer 1

The ice water mixture at the beginning and end is at 0 deg C

added x kg of ice

warm up x kg of ice from -15 to 0
Hi = heat into ice = Cice * x *15

That heat has to come from water freezing. Say y kg of water freezes
so
Cice * x * 15 = heat of fusion * y

Amount of ice at the end
so
.884 = .450 + y + x

You now have two equations in x and y. Solve for x :)

Answer 2

To find the mass of ice that was added, we can use the principle of conservation of energy.

First, let's find the heat gained by the water and ice in the bucket.
The heat gained by the water can be calculated using the specific heat capacity of water (c) and the change in temperature:
Q_water = m_water * c * ΔT_water

For the ice, we need to consider the latent heat of fusion:
Q_ice = m_ice * L_f

The total heat gained by the water and ice is equal to the heat lost by the added ice:
Q_water + Q_ice = m_added_ice * L_f

Since there is no heat exchange with the surroundings, the total heat gained by the water and ice is zero:
Q_water + Q_ice = 0

We can rearrange the equation to solve for the mass of added ice:
m_added_ice = -(Q_water + Q_ice) / L_f

Now we can plug in the given values and solve for the mass of added ice.

Given:
m_water = 1.75 kg
m_initial_ice = 0.450 kg
m_total_ice = 0.884 kg
T_initial_ice = -15.0 °C

The specific heat capacity of water, c, is approximately 4186 J/kg°C.
The latent heat of fusion, L_f, for water is approximately 334,000 J/kg.

First, let's calculate the change in temperature ΔT_water:
ΔT_water = 0 °C - (-15.0 °C) = 15.0 °C

Next, let's calculate the heat gained by the water:
Q_water = m_water * c * ΔT_water
Q_water = 1.75 kg * 4186 J/kg°C * 15.0 °C

Now, let's calculate the heat gained by the initial ice:
Q_ice = m_initial_ice * L_f

Now, let's calculate the mass of added ice:
m_added_ice = -(Q_water + Q_ice) / L_f

Finally, let's substitute the values into the equation and solve for m_added_ice.

Answer 3

To find the mass of ice that was added, we need to use the principle of conservation of energy. Since there is no heat exchange with the surroundings, we can assume that the heat lost by the water and ice is gained by the added ice.

First, let's find the heat lost by the water and ice in the bucket. We can use the formula:

Q = mcΔT

where Q is the heat lost, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

For the water, assuming its specific heat capacity is 4186 J/kg·K and its initial temperature is 0°C:

Q_water = (mass of water) x (specific heat capacity of water) x (change in temperature)

Q_water = 1.75 kg x 4186 J/kg·K x (0°C - 0°C) = 0 J

Since the water is at 0°C (the freezing point of water), it doesn't need to lose any heat to reach thermal equilibrium.

For the ice, assuming its specific heat capacity is 2093 J/kg·K and its initial temperature is 0°C:

Q_ice = (mass of ice) x (specific heat capacity of ice) x (change in temperature)

Q_ice = 0.450 kg x 2093 J/kg·K x (0°C - 0°C) = 0 J

Similarly, since the ice is at 0°C, it also doesn't need to lose any heat.

Now, the heat gained by the added ice can be calculated using the formula:

Q_added_ice = (mass of added ice) x (specific heat capacity of ice) x (change in temperature)

Since the added ice comes from a refrigerator at -15.0°C, the change in temperature is:

ΔT = (-15.0°C - 0°C) = -15.0°C

Now, we can rearrange the formula to solve for the mass of added ice:

mass of added ice = Q_added_ice / (specific heat capacity of ice x change in temperature)

mass of added ice = Q_added_ice / (2093 J/kg·K x -15.0°C)

Substituting Q_added_ice = Q_water + Q_ice = 0 J + 0 J = 0 J:

mass of added ice = 0 / (2093 J/kg·K x -15.0°C)

Note that the temperature difference ΔT is negative because heat is gaining in the case of added ice.

Therefore, the mass of added ice is 0 kg, indicating that no ice was added to reach thermal equilibrium.