Question:
Steam at a temperature of $100^{\circ}C$ is passed into water of mass 90 g such that the temperature of the water increases from $20^{\circ}C$ to $40^{\circ}C$ Then the total mass of the water at $40^{\circ}C$ is}
Steam at a temperature of $100^{\circ}C$ is passed into water of mass 90 g such that the temperature of the water increases from $20^{\circ}C$ to $40^{\circ}C$ Then the total mass of the water at $40^{\circ}C$ is}
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Always include latent heat when steam condenses.
Updated On: Oct 6, 2026
- 3 g
- 93 g
- 30 g
- 120 g \bigskip
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Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Concept:
Heat lost by steam = heat gained by water. Latent heat of steam plays key role.
Step 1: Heat gained by water.
\[ Q = mc\Delta T = 90 \times 1 \times (40-20) = 1800~cal \]
Step 2: Steam releases latent heat.
Let mass of steam = $m$: \[ m(540 + 60) = 1800 \] \[ 600m = 1800 \Rightarrow m = 3g \]
Step 3: Total mass.
\[ = 90 + 3 = 93g \] Final Answer: \[ (B)\ 93g \]
Step 1: Heat gained by water.
\[ Q = mc\Delta T = 90 \times 1 \times (40-20) = 1800~cal \]
Step 2: Steam releases latent heat.
Let mass of steam = $m$: \[ m(540 + 60) = 1800 \] \[ 600m = 1800 \Rightarrow m = 3g \]
Step 3: Total mass.
\[ = 90 + 3 = 93g \] Final Answer: \[ (B)\ 93g \]
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