NEET Physics Question - Thermal Physics

If there are no heat losses, the heat released by the condensation of x gm of steam at 100ºC into water at 100ºC can be used to convert y gm of ice at 0ºC into water at 100ºC. Then the ratio y : x is nearly :

1 : 1

2.5 : 1

2 : 1

3 : 1

Solution:

1. Heat released by \(x\) gm steam at \(100^\circ C\):
\[
Q_{\text{steam}} = x \times 540
\]

2. Heat required to convert \(y\) gm ice at \(0^\circ C\) to water at \(100^\circ C\):
\[
Q_{\text{ice}} = y \times 80 + y \times 100 = y \times 180
\]

3. Equating:
\[
x \times 540 = y \times 180
\]

\[
\frac{y}{x} = 3
\]

So, \( y : x = 3 : 1 \).

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Solution:

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