10-1.Thermometry, Thermal Expansion and Calorimetry
hard

$100°C$ 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?

A

$0.13$

B

$0.065$

C

$0.26$

D

$0.135$

(IIT-1995)

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

$560\, m = 1.12  \times  65, m = 0.130 \,gm$

Standard 11
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.