logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Ad
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
0 votes

60 gm of ice at $\;-5^{0}C\;$ is mixed with 100 gm of water at $\;50^{0}C\;$. Latent heat of fusion = 80 cal/g , Specific heat of water = $\;1 cal/g^{0}C\;$ ,Specific heat of ice = $\;0.5 cal/g^{0}C$ . Then the mass of ice left in the mixture = ?

$(a)\;12.5 \;gm\qquad(b)\;25 \;gm\qquad(c)\;60 \;gm\qquad(d)\;0 \;gm$

Can you answer this question?
 
 

1 Answer

0 votes
Answer : (d) $\;0\;gm$
Explanation :
Heat required to bring ice to $\;0^{0}C$
$=60 \times 10 - (-5) \times 0.5$
$=150 \;cal$
Heat required to melt ice = $60 \times 80$
$=4800 cal$
Max heat that can be lost by water
$=100 \times 1 \times 50$
$=5000 cal$
Since the total heat needed to melt all ice can be lost by water , mass of ice left = 0 g
answered Mar 12, 2014 by yamini.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...