A 68 g ice cube at −23◦C is dropped into a container of water at 0◦ C.
How much water freezes onto the ice? The specific heat of ice is 0.5 cal/g ·◦ C and its heat of fusion of is 80 cal/g.
Answer in units of g.
assume the final temp is 0C (we can check that later)
heat to melt ice+Heat to freeze water=0
68*specific heat of ice*(0- (-23)+Mass*Hf=0
mass= 68*.5*23/80=9.78 grams
check my thinking.
To determine how much water freezes onto the ice, we need to calculate the amount of heat transferred from the water to the ice.
First, let's calculate the heat required to raise the temperature of the ice cube to 0°C:
q1 = mass × specific heat × ΔT
= 68 g × 0.5 cal/g · °C × (0 - (-23)°C)
= 68 g × 0.5 cal/g · °C × 23°C
= 782 cal
Next, we need to determine the heat released during the phase change from ice to water at 0°C:
q2 = mass × heat of fusion
= 68 g × 80 cal/g
= 5440 cal
Now, let's calculate the total heat transferred from the water to the ice:
q_total = q1 + q2
= 782 cal + 5440 cal
= 6222 cal
Since 1 g of water releases 1 cal of heat when it freezes, we can calculate the amount of water that freezes onto the ice cube by dividing the total heat transferred by the heat released per gram:
mass_water_frozen = q_total / heat released per gram
= 6222 cal / 1 cal/g
= 6222 g
Therefore, approximately 6222 g of water freezes onto the ice cube.
To determine how much water freezes onto the ice cube, we need to calculate the heat transfer between the ice cube and the water.
The heat transfer equation can be written as:
Q = m × c × ΔT
where Q is the heat transfer, m is the mass, c is the specific heat, and ΔT is the change in temperature.
Step 1: Calculate the heat transfer required to raise the temperature of the ice from -23°C to 0°C.
Q1 = m_ice × c_ice × (0°C - (-23°C))
Substituting the values:
Q1 = (68 g) × (0.5 cal/g·°C) × (0°C - (-23°C))
Q1 = 68 g × 0.5 cal/g·°C × 23°C
Q1 = 782 cal
Step 2: Calculate the heat transfer required to change the state of the ice from a solid to a liquid.
Q2 = m_ice × ΔH_fusion
where ΔH_fusion is the heat of fusion.
Substituting the values:
Q2 = (68 g) × (80 cal/g)
Q2 = 5440 cal
Step 3: Calculate the total heat transfer required.
Total heat transfer (Q_total) = Q1 + Q2
Q_total = 782 cal + 5440 cal
Q_total = 6222 cal
Step 4: Calculate the mass of water that freezes onto the ice cube.
We will assume that all the heat transfer comes from the water, and none from the surroundings.
Q_total = m_water × c_water × ΔT
Since we want to find the mass of water that freezes, we rearrange the equation to solve for m_water:
m_water = Q_total / (c_water × ΔT)
The specific heat of water, c_water, is approximately 1 cal/g·°C.
Substituting the values:
m_water = 6222 cal / (1 cal/g·°C × (0°C - 0°C))
m_water = 6222 cal / (1 cal/g·°C × 0°C)
Since dividing by zero is undefined, we cannot determine the exact mass of water that freezes onto the ice cube. However, we can conclude that no water will freeze onto the ice cube because the water is already at 0°C, and there is no temperature difference to drive the heat transfer.