Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
+1 vote
87 views
in Physics by (90.2k points)
closed by
1 g of ice at `0^@C` is mixed with 1 g of steam at `100^@C`. After thermal equilibrium is achieved, the temperature of the mixture is
A. `50^(@)`C
B. `0^(@)`C
C. `55^(@)`C
D. `100^(@)`C

1 Answer

0 votes
by (90.9k points)
selected by
 
Best answer
Correct Answer - D
d) According to principle of calorimetry, total heat given by a hotter body is equal to the total heat received by colder body i.e., heat lost by hotter body = heat gained by colder body
Heat required to melt 1g of ice at `0^(@)`C into 1g of water at `0^(@)`C is 80 cal.
Heat required to convert 1g of water at `0^(@)`C into 1g of water at `100^(@)`C = `1 xx 1xx100 = 100 cal`
Heat required to condense 1g of steam = `1xx540` cal = 540 cal
Clearly, whole of steam is not condensed, So, temperature of the mixture is `100^(@)`C.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...