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$100°C$ તાપમાને રહેલ વરાળ $0.02 \,kg$ જળતુલ્યાંક ધરાવતા કેલરીમીટરમાં $15°C$ તાપમાને રહેલ $1.1\, kg$ પાણી પરથી પસાર થાય જ્યાં સુધી કેલરીમીટર અને પાણીનું તાપમાન $80°C$ થાય.તો કેટલા $kg$ વરાળનું પાણીમાં રૂપાંતર થયું હશે?
$0.13$
$0.065$
$0.26$
$0.135$
Solution
(a) Heat is lost by steam in two stages
$(i)$ for change of state from steam at $100°C$ to water at $100°C$ is $m \times 540$
$(ii)$ to change water at $100°C $ to water at $80°C$ is $m \times 1 \times (100 -80),$ where $m$ is the mass of steam condensed.
Total heat lost by steam is $m \times 540 + m \times 20 = 560\, m (cals)$ Heat gained by calorimeter and its contents is
$= (1.1 + 0.02) \times (80 -15) = 1.12 \times 65 \,cals.$
using Principle of calorimetery, Heat gained $=$ heat lost
$560\, m = 1.12 \times 65, m = 0.130 \,gm$