An experiment takes 10min to raise the temperature of water in a container from 0 ^ o C to 100 ^ o C and another 55min to convert it totally into steam by a heater supplying heat at a uniform rate. Neglecting the

Correct answer: B. 550cal g ^ - 1 Let the constant state of supply of heat per unit time = λ . So according to the first condition. λ t _ 1 = mc × 100 ⇒ (λ)/(m) = 100C t _ 1 . . . ( i ) , and according to the second condition, λ t _ 2 = m L _ f ⇒ (λ)/(m) = L _ f t _ 2 . . . ( ii ) . By ( i ) and ( ii ) 100C t _ 1 = L _ f t _ 2 ⇒ (1cal × 100)/(10) = L _ f 55 ⇒ L _ f = 550cal g ^ - 1

JEE Main 201511 AprilOnline

Exam: JEE Main
Year: 2015
Date: 11 April
Mode: Online
Subject: Physics
Chapter: Thermal Properties of Matter

JEE Main 2015 · 11 April · Online · Physics · Thermal Properties of Matter

An experiment takes 10min to raise the temperature of water in a...

An experiment takes

10min

to raise the temperature of water in a container from

{0}^{o}C

{100}^{o}C

and another

55min

to convert it totally into steam by a heater supplying heat at a uniform rate. Neglecting the specific heat of the container and taking specific heat of the water to be

1cal{\left({g}^{∘}C\right)}^{- 1}

, the heat of vaporization according to this experiment will come out to be:

Options

A $530cal{g}^{- 1}$
B $550cal{g}^{- 1}$
C $540cal{g}^{- 1}$
D $560cal{g}^{- 1}$

Correct answer

B. $550cal{g}^{- 1}$

Step-by-step solution

Let the constant state of supply of heat per unit time

= \lambda

. So according to the first condition.

\lambda{t}_{1} = mc \times 100 ⇒ \dfrac{\lambda}{m} = \dfrac{100C}{{t}_{1}} . . . ( i )

, and according to the second condition,

\lambda{t}_{2} = m{L}_{f} ⇒ \dfrac{\lambda}{m} = \dfrac{{L}_{f}}{{t}_{2}} . . . ( ii )

. By

( i )

and

( ii )

\dfrac{100C}{{t}_{1}} = \dfrac{{L}_{f}}{{t}_{2}}

⇒ \dfrac{1cal \times 100}{10} = \dfrac{{L}_{f}}{55} ⇒ {L}_{f} = 550cal{g}^{- 1}

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