Questions
4 gm of steam at is added to 20 gm of water at in a container of negligible mass. Assuming no heat is lost to surrounding, the mass of water in container at thermal equilibrium is. Latent heat of water in container 540 cal/gm. Specific heat of water = 1 cal/gm-
detailed solution
Correct option is C
Heat released by steam in conversion to water a 1000C is Q1 = mL = 4×540 = 2160 cal. Heat required to raise tempreature of water from 460C to 1000C is Q2 = ms∆θ = 20×1×54 = 1080J Q1 > Q2 and Q1Q2 = 2Hence all steam is not converted to water only half steam shall be converted to waterFinal mass of water = 20 + 2 = 22 gmTalk to our academic expert!
Similar Questions
We can get vapour state of a substance through the process of:
(i) boiling
(ii) evaporation
(iii) sublimation
(iv) regelation
799 666 8865
support@infinitylearn.com
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
JEE Mock Tests
JEE study guide
JEE Revision Notes
JEE Important Questions
JEE Sample Papers
JEE Previous Year's Papers
NEET previous year’s papers
NEET important questions
NEET sample papers
NEET revision notes
NEET study guide
NEET mock tests