A foam cooler of negligible mass contains 1.75 kg of water. Ice from a refrigerator at -15°C, is then added to the water in the

cooler. When thermal equilibrium has been reached, the total mass of ice in the cooler is 0.778 kg. Assuming no heat
exchange with the surroundings, what mass of ice was added? For water, L4 = 334 x 103 J/kg, and for ice, C = 2100 J/(kg .K).

To calculate the mass of ice that was added to the foam cooler, we can use the principle of energy conservation in a closed system.

The first step is to calculate the heat gained by the ice when it melts to reach the thermal equilibrium with water.

The heat gained by the ice can be calculated using the formula:
Q = m_ice * L_fusion

Where:
Q is the heat gained by the ice (in joules)
m_ice is the mass of ice that melted (in kilograms)
L_fusion is the latent heat of fusion for ice (in joules per kilogram)

The latent heat of fusion for ice is the amount of heat required to convert a unit mass of ice at its melting point to water at the same temperature. In this case, L_fusion = 334 x 10^3 J/kg.

We also know that the heat gained by the ice is equal to the heat lost by the water, so:

Q = m_water * C_water * ΔT

Where:
m_water is the mass of water (in kilograms)
C_water is the specific heat capacity of water (in joules per kilogram per degree Celsius)
ΔT is the change in temperature of the water (in degrees Celsius)

In this case, the change in temperature of the water is from the melting point of ice (0°C) to the final equilibrium temperature (unknown). We'll call this final temperature T.

Now we can set up an equation to solve for the mass of ice that melted:

m_ice * L_fusion = m_water * C_water * (T - 0)

Substituting the given values:
0.778 kg * 334 x 10^3 J/kg = 1.75 kg * 2100 J/(kg .K) * (T - 0)

Simplifying the equation:
258.932 = 3675 (T - 0)

Dividing both sides by 3675:
T - 0 = 0.07054

Therefore, the final equilibrium temperature T is approximately 0.07054 degrees Celsius.

To find the mass of ice that was added, we need to calculate the change in temperature of the ice:

ΔT_ice = T - (-15)
ΔT_ice = 0.07054 - (-15)
ΔT_ice = 15.07054 degrees Celsius

Now, we can use the equation for the specific heat capacity of ice:

Q = m_ice * C_ice * ΔT_ice

Where:
Q is the heat gained by the ice (in joules)
m_ice is the mass of ice (in kilograms)
C_ice is the specific heat capacity of ice (in joules per kilogram per degree Celsius)

In this case, the specific heat capacity of ice is known to be 2100 J/(kg . K).

Since the ice reached the melting point, there is no change in phase, so no latent heat of fusion is involved.

Therefore, the heat gained by the ice can be calculated as follows:

Q = m_ice * C_ice * ΔT_ice
Q = m_ice * 2100 J/(kg . K) * 15.07054 degrees Celsius

We know that this heat gained is the same as the heat lost by the water, which was calculated earlier as 258.932 J.

Therefore, we can set up the equation:

258.932 J = m_ice * 2100 J/(kg . K) * 15.07054 degrees Celsius

Dividing both sides by 2100 * 15.07054 J/(kg . K):
m_ice = 258.932 J / (2100 J/(kg . K) * 15.07054 degrees Celsius)

Now, we can calculate the mass of ice that was added:

m_ice = 0.778 kg * 258.932 J / (2100 J/(kg . K) * 15.07054 degrees Celsius)

After doing the calculations, the mass of ice that was added is approximately 0.06045 kg or 60.45 g.