Two 20.0g ice cubes at -10.0 C are placed into 275g 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 to melt ice cube 1 + heat t melt ice cube 2 + heat to raise T ice melted ice cubes from zero to final T + heat lost by 275 g H2O = 0.
heat to melt ice cube 1 = mass x heat fusion.
heat t melt ice cube 2 = mass x heat fusion.
heat to raise water from 20 g ice cubes from zero to final T = mass x specific heat x (Tfinal-Tintial)
heat lost by 275 g water = mass water x specific heat water x (Tfinal-Tintial) = zero. Solve for Tfinal. I something a little less than 20 C.
To solve this problem, we need to apply the principle of conservation of energy. The heat lost by the water when it cools down must be equal to the heat gained by the ice when it melts, as there is no transfer of energy to or from the surroundings.
Let's break down the problem into two stages:
Stage 1: The ice cubes are heated to reach the melting point (0°C).
Stage 2: The ice melts to water, and the water reaches a final temperature.
Stage 1: To heat the ice cubes from -10°C to 0°C, we can use the specific heat capacity formula:
Q = mcΔT
Where:
Q = Heat transferred (in joules)
m = Mass of the ice cubes
c = Specific heat capacity of ice (2.093 J/g°C)
ΔT = Change in temperature
Using the given values:
m = 20.0g (mass of each ice cube)
c = 2.093 J/g°C
ΔT = (0°C) - (-10°C) = 10°C
Q1 = (20.0g)(2.093 J/g°C)(10°C) = 418.6 J (joules)
Stage 2: Now that the ice has reached the melting point, the heat gained by the ice to melt it should be equal to the heat lost by the water as it cools down from 25°C to the final temperature.
To calculate the final temperature, we can use the formula:
Q2 = mcΔT
Where:
Q2 = Heat transferred (in joules)
m = Mass of the water
c = Specific heat capacity of water (4.186 J/g°C)
ΔT = Change in temperature (final temperature - initial temperature)
Using the given values:
m = 275g (mass of water)
c = 4.186 J/g°C
ΔT = Tf - 25°C
Q2 = (275g)(4.186 J/g°C)(Tf - 25°C)
Since the heat lost in stage 2 is equal to the heat gained in stage 1, we can set up an equation:
Q1 = Q2
418.6 J = (275g)(4.186 J/g°C)(Tf - 25°C)
Now we can solve for the final temperature, Tf:
418.6 J = (275g)(4.186 J/g°C)(Tf - 25°C)
Divide both sides by (275g)(4.186 J/g°C):
418.6 J / (275g)(4.186 J/g°C) = (Tf - 25°C)
1.2207 = Tf - 25°C
Tf = 26.2207°C
Therefore, the final temperature of the water after all the ice melts would be approximately 26.2°C.