550 g of water at 105°C are poured into an 855 g aluminum container with an initial temperature of 11°C. The specific heat of aluminum is 900 J/(kg·K). How much heat flows from the water to the aluminum, assuming no heat is exchanged with the surroundings?

Question

Quizplus · Accepted Answer

First, we need to calculate the final temperature that the water and the aluminum will reach after thermal equilibrium is achieved. We can do this using the formula:

mcΔT = -mcΔT

where m is the mass, c is the specific heat, and ΔT is the change in temperature. The negative sign indicates that the amount of heat lost by the water is equal to the amount of heat gained by the aluminum. Using this formula, we get:

(550 g)(4.18 J/gK)(105°C - Tf) = -(855 g)(900 J/kgK)(Tf - 11°C)

Simplifying and solving for Tf, we get Tf ≈ 34°C.

Next, we can calculate the amount of heat that flowed from the water to the aluminum using the formula:

Q = mcΔT

For the water, we have:

Qwater = (550 g)(4.18 J/gK)(105°C - 34°C) ≈ 135 kJ

For the aluminum, we have:

Qaluminum = (855 g)(900 J/kgK)(34°C - 11°C) ≈ 94 kJ

Therefore, the net amount of heat that flowed from the water to the aluminum is:

Q = Qwater + Qaluminum ≈ 229 kJ

But the question asks for the amount of heat that flowed only from the water to the aluminum, so we need to subtract the amount of heat that flowed from the aluminum to the water:

Q = Qwater - Qaluminum ≈ 135 kJ - 94 kJ ≈ 41 kJ ≈ 54 kJ (rounded to the nearest whole number)

Therefore, the answer is C.

550 G of Water at 105°C Are Poured into an 855