A 100 gram aluminum container is at 15.0$^{\circ}$C.150 grams of water at 80.0$^{\circ}$C is added to the aluminum container.What is the final temperature of the water and the aluminum? (specific heat of water = 1.00 cal/g$^{\circ}$C,specific heat of aluminum = 0.215 cal/gm$^{\circ}$C)

Question

Quizplus · Accepted Answer

First, we need to calculate the heat lost by the aluminum container and the heat gained by water:

Q(aluminum) = m(aluminum) * C(aluminum) * ΔT
Q(water) = m(water) * C(water) * ΔT

Where m is the mass, C is the specific heat, and ΔT is the change in temperature.

The heat lost by the aluminum container is equal to the heat gained by the water, so:

Q(aluminum) = -Q(water)

m(aluminum) * C(aluminum) * ΔT = -m(water) * C(water) * ΔT

Solving for ΔT:

ΔT = -m(water) * C(water) * ΔT / (m(aluminum) * C(aluminum) )

Substituting the values:

ΔT = -150 g * 1.00 cal/g °F * (80.0 °F - T) / (100 g * 0.215 cal/gm °F)

where T is the final temperature of both the water and the aluminum container.

Solving for T:

T = 68.5 °F

Therefore, the final temperature of both the water and the aluminum container is 68.5°F. The answer is choice C.

A 100 Gram Aluminum Container Is at 15 ∘^{\circ}∘ C150 Grams of Water at 80

A 100 Gram Aluminum Container Is at 15 $^{\circ}$ C150 Grams of Water at 80