Two 20.0-g ice cubes at –20.0 °C are placed into 295 g of water at 25.0 °C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature of the water after all the ice melts.
Heat capacity of H2O(s)=37.7J
Heat capacity of H2O(l)=75.3J
Enthalpy of fusion of H2)=6.01kJ
To find the final temperature of the water after all the ice melts, we need to calculate the heat transferred between the ice and the water.
First, let's calculate the heat released by the two ice cubes when they melt:
Heat released by the first ice cube = mass of ice cube 1 × heat capacity of H2O(s) × temperature change of H2O(s)
= 20.0 g × 37.7 J/g °C × (0 °C - (-20.0 °C))
= 20.0 g × 37.7 J/g °C × 20.0 °C
= 15,080 J
Heat released by the second ice cube = mass of ice cube 2 × heat capacity of H2O(s) × temperature change of H2O(s)
= 20.0 g × 37.7 J/g °C × (0 °C - (-20.0 °C))
= 20.0 g × 37.7 J/g °C × 20.0 °C
= 15,080 J
The total heat released by melting the two ice cubes is 15,080 J + 15,080 J = 30,160 J.
Next, let's calculate the heat absorbed by the water to raise its temperature:
Heat absorbed by the water = mass of water × heat capacity of H2O(l) × temperature change of water
= 295 g × 75.3 J/g °C × (final temperature - 25.0 °C)
Since we know that no energy is transferred to or from the surroundings, the heat released by the ice must be equal to the heat absorbed by the water:
30,160 J = 295 g × 75.3 J/g °C × (final temperature - 25.0 °C)
Let's solve this equation for the final temperature:
30,160 J = 22,193.5 J/°C × (final temperature - 25.0 °C)
(final temperature - 25.0 °C) = 30,160 J / 22,193.5 J/°C
(final temperature - 25.0 °C) = 1.359 °C
final temperature = 25.0 °C + 1.359 °C
final temperature ≈ 26.359 °C
Therefore, the final temperature of the water after all the ice melts is approximately 26.359 °C.
To calculate the final temperature of the water after all the ice melts, we need to consider the heat transfer between the ice and the water. We can do this by applying the principle of conservation of energy.
First, let's calculate the heat required to melt the ice. The heat required to melt one ice cube can be calculated using the formula:
Heat = mass × enthalpy of fusion
Given:
Mass of one ice cube = 20.0 g
Enthalpy of fusion of H2O = 6.01 kJ
Converting the mass of the ice cube to kilograms:
Mass of one ice cube = 20.0 g = 0.02 kg
Calculating the heat required to melt one ice cube:
Heat = 0.02 kg × 6.01 kJ/kg = 0.1202 kJ
Now, let's calculate the heat transferred from the water to the ice. The heat transferred can be calculated using the formula:
Heat = mass × specific heat capacity × change in temperature
Given:
Mass of water = 295 g
Specific heat capacity of water (in liquid form) = 75.3 J/g°C
Initial temperature of the water = 25.0°C
Converting the mass of water to kilograms:
Mass of water = 295 g = 0.295 kg
Calculating the initial heat of the water:
Heat = 0.295 kg × 75.3 J/g°C × (0.0°C - 25.0°C)
Heat = -554.925 J
Now, let's calculate the change in temperature of the ice melter, assuming it started at -20.0°C and ends up at 0.0°C when it melts:
Change in temperature = 0.0°C - (-20.0°C) = 20.0°C
Next, let's consider the heat gained by the water as it heats up from 0.0°C to its final temperature:
Heat = mass × specific heat capacity × change in temperature
Given:
Specific heat capacity of water (in solid form) = 37.7 J/g°C
Mass of water = 295 g
Final temperature of the water = ?
Converting the mass of water to kilograms:
Mass of water = 295 g = 0.295 kg
Let's calculate the heat gained by the water:
Heat = 0.295 kg × 37.7 J/g°C × (Tf - 0.0°C), where Tf is the final temperature of the water
Now, we can set up the conservation of energy equation:
Heat gained by the water = Heat lost by the ice
0.295 kg × 37.7 J/g°C × (Tf - 0.0°C) = -554.925 J + 0.1202 kJ
Simplifying the equation:
0.295 × 37.7 × Tf = -554.925 + 0.1202
11.0165 × Tf = -554.925 + 0.1202
11.0165 × Tf = -554.8048
Finally, we can solve for the final temperature of the water (Tf):
Tf = (-554.8048) / 11.0165
Calculating Tf:
Tf ≈ -50.333 °C
Therefore, the final temperature of the water after all the ice melts is approximately -50.333 °C.
heat to raise 40 g ice from -20 to zero.
q1 = mass x specific heat x (Tfinal-Tinitial)
heat to melt 40 g ice.
q2 = mass ice x heat fusion (in joules)
heat to raise melted water to final T.
q3 = mass water x specific liquid water x (Tfinal-Tinitial)
heat absorbed by 295 g H2O @ 25.0C
q4 = mass H2O x specific heat liquid H2O x (Tfinal-Tinitial).
q1 + q2 + q3 + q4 = 0
Substitute and solve for Tfinal. Watch the signs.