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What weight of ice is melted at 0°C by the heat liberated by condensing 180 g of super heated steam at 150°C (b.p of water=100°C, heat of vaporization =540 cal g¹, heat of fusion of ice =80 cal g¹, specific heat od steam=1.6 cal g¹ degree ¹.​

User Tchaka
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Answer: 1395 grams of ice can be melted

Step-by-step explanation:

To find the weight of ice that is melted by the heat liberated by condensing 180 g of superheated steam at 150°C, we need to use the following information:

The boiling point of water is 100°C

The heat of vaporization of water is 540 cal/g

The heat of fusion of ice is 80 cal/g

The specific heat of steam is 1.6 cal/g°C

First, we need to calculate the total heat energy released by the condensation of the steam. This can be found by using the heat of vaporization and the amount of steam being condensed:

180 g x 540 cal/g = 97200 cal

Next, we need to calculate the heat energy absorbed by the steam as it cools down from 150°C to 100°C. This can be found using the specific heat of steam and the change in temperature:

180 g x 1.6 cal/g°C x (100 - 150) = -14400 cal

Now, we can subtract the heat absorbed by the steam from the total heat released by the condensation to find the heat energy available to melt the ice:

97200 cal - (-14400 cal) = 111600 cal

Finally, we can divide this heat energy by the heat of fusion of ice to find the weight of ice that can be melted:

111600 cal / 80 cal/g = 1395 g

Therefore, 1395 grams of ice can be melted

User The Techel
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