5.5k views
1 vote
A drop of water weighing 0.48 g is vaporized at 100 ?c and condenses on the surface of a 55-g block of aluminum that is initially at 25 ?c. if the heat released during condensation goes only toward heating the metal, what is the final temperature in celsius of the metal block? (the specific heat capacity of aluminum is 0.903 j/g ?c.

2 Answers

3 votes

Final answer:

The question involves calculating the final temperature of an aluminum block after it absorbs heat from the condensation of a drop of water. Using the conservation of energy principle and specific heat capacity, the temperature change of the aluminum block can be determined by equating the heat released by the water to the heat absorbed by the aluminum.

Step-by-step explanation:

Calculating the Final Temperature of an Aluminum Block

The student's question involves a thermal energy transfer process where a drop of water vaporizes and condenses on an aluminum block, releasing heat. To find the final temperature of the aluminum block, we need to use the principle of conservation of energy, which states that the heat lost by the water during condensation must be equal to the heat gained by the aluminum block. We can use the formula:

Q = m × c × ΔT

Where:

  • Q is the heat transferred
  • m is the mass
  • c is the specific heat capacity
  • ΔT is the change in temperature

The heat released during condensation of water at 100°C (Qwater) can be calculated using the latent heat of vaporization of water (2260 J/g). Since the mass of the water drop is 0.48 g, we have:

Qwater = mwater × latent heat of vaporization

Qwater = 0.48 g × 2260 J/g

This energy will be absorbed by the aluminum block. The heat gained by the aluminum block (Qaluminum) is:

Qaluminum = maluminum × caluminum × (ΔT)

Since Qwater = Qaluminum, we solve for ΔT.

Finally, we add the ΔT to the initial temperature of the aluminum block to find the final temperature of the metal block.

User Gardener
by
7.3k points
5 votes

Hi, the solution would be like this for this specific problem:

(2257 J/g) x (0.48 g) = 1083.36 J

(1083.36 J) / (0.903 J/g ∘C) / (55 g) = 21.81°C change


25°C + 21.81°C = 47°C


I hope this helps and if you have any further questions, please don’t hesitate to ask again.

User Reynaldo Aguilar
by
7.8k points