Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

A sample of 0.1 g of water at l00°C and normal pressure.(1.013 x 105 Nm-2 )requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample, is:

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

a

104.3 J

b

208.7 J

c

42.2 J

d

84.5 J

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

Q=54 cal =54×4.18 joule=225.72 joule W=PVsteam-VwaterFor water 0.1 gram = 0.1 cc =0.013×105167.1×10-6-0.1×10-6joule =1.013×167×10-1=16.917 joule By FLOT ⇒  ∆U=Q-W=225.72-16.917=208.8 joule.
Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring
A sample of 0.1 g of water at l00°C and normal pressure.(1.013 x 105 Nm-2 )requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample, is: