0.115 kg of water at 81.0o C is poured into an insulated cup containing 0.220 kg of ice initally at 0o. Calculate the mass of liquid when the system reaches thermal equilibrium.
In this case, not all of the ice melts. Let m be the number of grams of ice that melts.
115*81*1.0 = m*80
(80 cal/g is the heat of fusion)
m = 116 g
Note that is less than the amount of ice you started out with. So there is some ice left.
Add that to the 115 g of liquid water you had at the beginning.
To solve this problem, we need to consider the heat transfer that occurs when the hot water cools down and melts the ice until both substances reach the same temperature and thermal equilibrium.
First, we need to calculate the amount of heat gained by the water as it cools down. We can use the formula:
Q = m * c * ΔT
Where:
Q = heat gained or lost
m = mass of the substance
c = specific heat capacity
ΔT = change in temperature
For water, the specific heat capacity (c) is approximately 4.18 J/g°C.
The water is initially at 81.0°C and cools down to the final temperature at which it reaches thermal equilibrium with the ice. Let's call this final temperature "T".
Now, let's calculate the heat gained by the water:
Q_water = (0.115 kg) * (4.18 J/g°C) * (T - 81.0°C)
Next, we need to calculate the heat lost by the ice as it melts. The heat required to melt ice is given by:
Q_ice = m * ΔH_fusion
Where:
ΔH_fusion = heat of fusion
For water, ΔH_fusion is approximately 334 J/g.
The ice is initially at 0°C and melts until it reaches the final temperature T. The mass of the ice is given as 0.220 kg.
Now, let's calculate the heat lost by the ice:
Q_ice = (0.220 kg) * (334 J/g) + (0.220 kg) * (4.18 J/g°C) * T
Since the system is insulated, the heat gained by the water is equal to the heat lost by the ice:
(0.115 kg) * (4.18 J/g°C) * (T - 81.0°C) = (0.220 kg) * (334 J/g) + (0.220 kg) * (4.18 J/g°C) * T
Now, we can solve for the final temperature (T):
(0.115 kg) * (4.18 J/g°C) * T - (0.115 kg) * (4.18 J/g°C) * 81.0°C = (0.220 kg) * (334 J/g) + (0.220 kg) * (4.18 J/g°C) * T
Simplifying and rearranging the equation:
(0.115 kg) * (4.18 J/g°C - 4.18 J/g°C) * T = (0.220 kg) * (334 J/g) + (0.115 kg) * (4.18 J/g°C) * (81.0°C)
(0.115 kg) * (4.18 J/g°C) * T = (0.220 kg) * (334 J/g) + (0.115 kg) * (4.18 J/g°C) * (81.0°C)
0.4787 kg * T = 73.48 J + 38.2822 kg * °C
T = (73.48 J + 38.2822 kg * °C) / 0.4787 kg
T = 216.55°C
The final temperature when the system reaches thermal equilibrium is 216.55°C.
To calculate the mass of liquid, we need to check the phase of the water at this final temperature. At temperatures above 100°C, water is in the gaseous phase (steam). Since the question asks for the mass of liquid, we need to subtract the mass of evaporated steam.
Given that the initial mass of water was 0.115 kg, and the final temperature is above 100°C, we can assume that all the water has evaporated.
Therefore, the mass of liquid when the system reaches thermal equilibrium is 0 kg.