A 250g aluminum cup holds and is in thermal equilibrium with 850g of water at 83 degrees celcuis. The combination of cup and water is cooled uniformly so that the temperature decreases by 1.5 degrees celcuis per minute. At what rate is energy being removed?
To find the rate at which energy is being removed, we can use the formula for heat transfer:
Q = mcΔT,
where Q is the heat transfer, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
First, let's calculate the heat transfer for the aluminum cup. The specific heat capacity of aluminum is 0.897 J/g°C. The mass of the cup is 250g, and the change in temperature is 1.5 degrees per minute. Therefore:
Q_cup = mcΔT
= 250g × 0.897 J/g°C × 1.5°C/min
Simplifying the equation:
Q_cup = 337.125 J/min
Next, let's calculate the heat transfer for the water. The specific heat capacity of water is 4.186 J/g°C. The mass of the water is 850g, and the change in temperature is also 1.5 degrees per minute. Therefore:
Q_water = mcΔT
= 850g × 4.186 J/g°C × 1.5°C/min
Simplifying the equation:
Q_water = 5708.35 J/min
Finally, to find the total rate at which energy is being removed, we add the heat transfers:
Q_total = Q_cup + Q_water
= 337.125 J/min + 5708.35 J/min
Simplifying the equation:
Q_total ≈ 6045.475 J/min
So, the rate at which energy is being removed is approximately 6045.475 J/min.