CalorimetryMedium
Question
2 gms steam at 100o C is mixed with 5 gm ice at - 40oC in an ideal calorimeter. The final temperature of the system will be (Given Lv = 500 cal/gm; Sice = 0.5 cal/gm oC, Sw = 1 cal/gm oC; Lf = 80 cal/gm) :
Options
A.0oC
B.100oC
C.80oC
D.50oC
Solution
Heat taken by ice to convert to water at 100oC fully
(5 × 40 × 0.5) + (5 × 80) + (5 × 100 × 1) = 100 cal.
Heat givenby steam
2 × 500 = 1000 cal. so steam and water both are at 100oC.
(5 × 40 × 0.5) + (5 × 80) + (5 × 100 × 1) = 100 cal.
Heat givenby steam
2 × 500 = 1000 cal. so steam and water both are at 100oC.
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