A 5-g ice cube at ?4°C is dropped into 60 g of water at 40°C. (For the following problems, use the specific heats in this table and the numerical data found in this table.)
(a) After enough time has passed to allow the ice cube and water to come into equilibrium, what is the temperature of the water?
°C
(b) If a second ice cube is added, what will the temperature be?
°C
Tf= 29.30 degree celsius
To solve this problem, we need to use the equation for heat transfer:
Q = m * c * ΔT
Where:
Q is the heat transferred
m is the mass of the substance
c is the specific heat of the substance
ΔT is the change in temperature
(a) To find the final temperature of the water after the ice cube and water come into equilibrium, we need to consider the heat transfer from the water to the ice cube until they reach the same temperature.
First, let's calculate the heat transfer from the water to reach its final temperature. Assuming the water's specific heat is 4.18 J/g°C:
Q_water = m_water * c_water * ΔT_water
m_water = 60 g (given)
c_water = 4.18 J/g°C (specific heat of water)
ΔT_water = final temperature - initial temperature = final temperature - 40°C
Next, let's calculate the heat transfer from the ice cube to reach the same final temperature. Assuming the ice cube's specific heat is 2.09 J/g°C:
Q_ice = m_ice * c_ice * ΔT_ice
m_ice = 5 g (given)
c_ice = 2.09 J/g°C (specific heat of ice)
ΔT_ice = final temperature - initial temperature = final temperature - (-4°C)
Since the heat lost by the water is equal to the heat gained by the ice cube, we can set the two equations equal to each other:
m_water * c_water * ΔT_water = m_ice * c_ice * ΔT_ice
Simplify the equation:
60 * 4.18 * (final temperature - 40) = 5 * 2.09 * (final temperature - (-4))
Now solve for the final temperature by isolating it:
250.8 * (final temperature - 40) = 10.45 * (final temperature + 4)
Simplify again:
250.8 * final temperature - 10,032 = 10.45 * final temperature + 41.8
Combine like terms:
250.8 * final temperature - 10.45 * final temperature = 41.8 + 10,032
240.35 * final temperature = 10,073.8
final temperature = 10,073.8 / 240.35
final temperature ≈ 41.9°C
Therefore, after the ice cube and water come into equilibrium, the temperature of the water is approximately 41.9°C.
(b) Now let's consider what happens when a second ice cube is added to the system. Since the water is already at its maximum temperature (41.9°C), the heat transfer will now occur from the water to melt the ice cube, rather than changing the temperature of the water.
To find the quantity of heat required to melt the ice, we need to use the equation:
Q_melt = m_ice * ΔH_fusion
Where ΔH_fusion is the heat of fusion for ice, which is 334 J/g.
m_ice = 5 g (given)
ΔH_fusion = 334 J/g
Q_melt = 5 * 334
Q_melt = 1670 J
Now, let's consider the quantity of heat transferred from the water to melt the ice cube. Assuming the specific heat of water is still 4.18 J/g°C, we can use the equation:
Q_water_melt = m_water * c_water * ΔT_water_melt
m_water = 60 g (given)
c_water = 4.18 J/g°C (specific heat of water)
ΔT_water_melt = final temperature - 41.9°C
Since the heat lost by the water is equal to the heat gained by the ice cube (to melt it):
Q_water_melt = Q_melt
Substituting the values:
60 * 4.18 * (final temperature - 41.9) = 1670
249.36 * (final temperature - 41.9) = 1670
249.36 * final temperature - 249.36 * 41.9 = 1670
249.36 * final temperature = 1670 + 10,438.784
249.36 * final temperature ≈ 12,108.784
final temperature ≈ 12,108.784 / 249.36
final temperature ≈ 48.6°C
Therefore, if a second ice cube is added, the final temperature of the water will be approximately 48.6°C.