Thermal Properties of Matter - Result Question 40
41. If $1 g$ of steam is mixed with $1 g$ of ice, then the resultant temperature of the mixture is
(a) $270^{\circ} C$
(b) $230^{\circ} C$
(c) $100^{\circ} C$
(d) $50^{\circ} C$
[1999]
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Answer:
Correct Answer: 41. (c)
Solution:
- (c) Heat required by ice at $0^{\circ} C$ to reach a temperature of $100^{\circ} C=m L+m c \Delta \theta$
$=1 \times 80+1 \times 1 \times(100-0)=180 cal$
Heat available with $1 g$ steam to condense into $1 g$ of water at $100^{\circ} C=536 cal$.
Obviously the whole steam will not be condensed and ice will attain a temperature of $100^{\circ} C$; so the temperature of mixture $=100^{\circ} C$.
