Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

When $M_{1}$ gram of ice at $-10^{\circ}C$ ( specific heat = $0.5$ cal $g^{-1}^{\circ}C^{-1}$ ) is added to $M_{2}$ gram of water at $50^{\circ}C$ , finally no ice is left and the water is at $0^{\circ}C$ . The value of latent heat of ice, in cal $g^{-1}$ is :

Option 1)
$\frac{5M_{1}}{M_{2}}-50$

Option 2)
$\frac{50M_{2}}{M_{1}}$

Option 3)
$\frac{5M_{2}}{M_{1}}-5$

Option 4)
$\frac{50M_{2}}{M_{1}}-5$

Answers (1)

best_answer

Latent Heat -

Q = mL

- wherein

Q = heat supplied

m = mass of substance

L = latent heat

As we know that

heat taken by ice = heat given by water

$M_{1}$ $S _{ice}$ $(10)+M_{1}Lg$ = $M_{2}$ $S _{water}$ $(50)$

$\Rightarrow \frac{M_{1}}{2}\times10+M_{1}Lg=M_{2}\times50$

$\Rightarrow 5+Lg=50\frac{M_{2}}{M_{1}}$

$\Rightarrow Lg=\frac{50M_{2}}{M_{1}}-5$

Option 1)

$\frac{5M_{1}}{M_{2}}-50$

Option 2)

$\frac{50M_{2}}{M_{1}}$

Option 3)

$\frac{5M_{2}}{M_{1}}-5$

Option 4)

$\frac{50M_{2}}{M_{1}}-5$

Posted by

solutionqc

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024 session 2 final answer key soon at jeemain.nta.ac.in; result on April 25

anam-khan

JEE Main 2024 Session 2: Last day to challenge answer key today

anam-khan

BTech admission 2024 without JEE; top government engineering colleges, upcoming entrance exams

Latest Question