Answer:
74.54g
Step-by-step explanation:
First, let’s calculate the heat lost by the tea when it cools down from 33.1°C to 26.3°C. We can use the formula Q = mcΔT, where Q is the heat lost, m is the mass of the tea, c is the specific heat capacity of water, and ΔT is the change in temperature.
In the context of the problem,
- m represents the mass of the tea, which is given as 185g.
- c represents the specific heat capacity of water, which is given as 4186 J/kg·°C.
- ΔT represents the change in temperature, which is (33.1 - 26.3)°C = 6.8°C.
So, the heat lost by the tea is: Q = (185g) * (4186 J/kg·°C) * (6.8°C) = 5345.68 J
This heat is gained by the ice, causing it to melt and warm up to 0°C. The heat required to melt ice is given by Q = mL, where m is the mass of ice melted and L is the latent heat of fusion of water.
The latent heat of fusion of water is 334 kJ/kg. So, we can calculate the mass of ice melted as: m = Q / L = (5345.68 J) / (334 kJ/kg) = 0.016 kg or 16g
However, not all of the ice will melt. Some of it will remain as ice and some will become water at 0°C.
Let’s say that x grams of ice melts completely and becomes water at 0°C. The remaining (90.3 - x) grams of ice will stay as ice.
The heat required to melt x grams of ice is: Q1 = x * L
The heat required to warm up x grams of water from 0°C to 26.3°C is: Q2 = x * c * (26.3 - 0)
The total heat gained by the ice and water is: Q = Q1 + Q2
Substituting the values we get: 5345.68 J = x * L + x * c * (26.3 - 0)
Solving for x, we get: x = (5345.68 J) / (L + c * (26.3 - 0)) ≈ 15.76g
Therefore, out of the initial 90.3g of ice, only approximately 15.76g melts completely and becomes water at 0°C.
The remaining mass of ice in the jar is: (90.3 - 15.76)g ≈ 74.54g.