Answer:
130.43g
Step-by-step explanation:
First, we need to calculate the heat lost by the tea when it cools down from 33.3°C to 31.8°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.
The mass of the tea is given as 187g and the specific heat capacity of water is 4186 J/kg·°C. The change in temperature is (33.3 - 31.8)°C = 1.5°C.
So, the heat lost by the tea is: Q = (187g) * (4186 J/kg·°C) * (1.5°C) = 1174.05 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 = (1174.05 J) / (334 kJ/kg) = 0.0035 kg = 3.5g
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 (133 - 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 31.8°C is: Q2 = x * c * (31.8 - 0)
The total heat gained by the ice and water is: Q = Q1 + Q2
Substituting the values we get: 1174.05 J = x * L + x * c * (31.8 - 0)
Solving for x, we get: x = 1174.05 J / (L + c * (31.8 - 0)) = 2.57g
Therefore, out of the initial 133g of ice, only 2.57g melts completely and becomes water at 0°C.
The remaining mass of ice in the jar is: 133g - 2.57g = 130.43g
So, at the time at which the temperature of the tea is 31.8°C, there are approximately 130.43g of ice remaining in the jar.