A 150.0 g aluminum cup is filled with 450. g of water at 95℃. The combination cools uniformly by 1.20℃ per minute. At what rate (J/s) is thermal energy being removed from the combination?

To find the rate at which thermal energy is being removed from the combination, we need to calculate the heat transfer per unit time. This can be done using the formula:

Q/t = -k * A * ΔT/Δt

Where:
Q/t is the rate of heat transfer
k is the thermal conductivity
A is the surface area of the cup
ΔT is the change in temperature
Δt is the change in time

First, we need to find the value of k for aluminum. The thermal conductivity of aluminum is approximately 205 W/(m·K).

Next, we need to calculate the surface area of the cup. Assuming the cup has a cylindrical shape, we can use the formula:

A = 2πrh + πr^2

Where:
A is the surface area
r is the radius of the cup
h is the height of the cup

Let's assume that the cup has a radius of 5 cm and a height of 10 cm. Converting the units to meters, we have r = 0.05 m and h = 0.1 m.

Therefore, the surface area of the cup A = 2π(0.05)(0.1) + π(0.05)^2 ≈ 0.157 m^2.

Now, we can calculate the change in temperature ΔT. The water starts at 95℃ and cools uniformly by 1.20℃ per minute. Let's say we observe the system after t minutes, then the change in temperature is given by:

ΔT = 95℃ - (95℃ + 1.20℃/min * t) = -1.20℃/min * t

Finally, we can plug in all the values into the formula to find the rate of heat transfer Q/t:

Q/t = -k * A * ΔT/Δt
= -(205 W/(m·K)) * (0.157 m^2) * (-1.20℃/min * t) / (1 min)
= 38.555 W * t

Therefore, the rate at which thermal energy is being removed from the combination is given by 38.555 watts per minute, or 38.555 J/s.