Answered

At Westonci.ca, we connect you with the best answers from a community of experienced and knowledgeable individuals. Join our Q&A platform to connect with experts dedicated to providing precise answers to your questions in different areas. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

A student attempted to measure the specific latent heat of vaporisation of water.

She took the following readings:

First balance reading (g) = 581
Second balance reading (g) = 526
First joulemeter reading (kJ) = 195
Second joulemeter reading (kJ) = 327

Use these results to determine the specific latent heat of vaporisation of water.

Sagot :

Answer:

The latent heat of vaporization of water is 2.4 kJ/g

Explanation:

The given readings are;

The first (mass) balance reading (of the water) in grams, m₁ = 581 g

The second (mass) balance reading (of the water) in grams, m₂ = 526 g

The first joulemeter reading in kilojoules (kJ), Q₁ = 195 kJ

The second joulemeter reading in kilojoules (kJ), Q₂ = 327 kJ

The latent heat of vaporization = The heat required to evaporate a given mass water at constant temperature

Based on the measurements, we have;

The latent heat of vaporization = ΔQ/Δm

∴ The latent heat of vaporization of water = (327 kJ - 195 kJ)/(581 g - 526 g) = 2.4 kJ/g

The latent heat of vaporization of water = 2.4 kJ/g

We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.