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In this scenario, the heat released by the condensation of \( m_1 \) grams of steam at 100°C into water at 100°C can be used to convert \( m_2 \) grams of ice at 0°C into water at 0°C, assuming no heat is lost to the surroundings. This process involves the transfer of heat without any loss, making it an idealized scenario for energy conservation. ✨

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Answer:

Yes, that statement is correct. If there is no heat loss to the surroundings, the heat released by the condensation of m1 grams of steam at 100°C can indeed be used to convert m2 grams of ice at 0°C into water at 0°C.

To understand this process, we need to consider the heat transfer involved. When steam condenses into water at its boiling point, it releases a certain amount of heat known as the heat of condensation. This heat can be used to increase the temperature of other substances or facilitate phase changes.

In this case, the heat released by the condensation of m1 grams of steam at 100°C is given by:

Q1 = m1 * heat of condensation

This heat is then used to melt m2 grams of ice at 0°C, which requires a certain amount of heat called the heat of fusion. The heat required to convert m2 grams of ice to water at 0°C is given by:

Q2 = m2 * heat of fusion

If there is no heat loss to the surroundings, Q1 must be equal to Q2 to ensure energy conservation. Therefore, we can equate the two equations:

m1 * heat of condensation = m2 * heat of fusion

By rearranging the equation, we can calculate the mass of steam (m1) required to melt a certain mass of ice (m2):

m1 = (m2 * heat of fusion) / heat of condensation

So, if there is no heat loss to the surroundings, the heat released by the condensation of steam can indeed be used to convert ice into water.