Let the constant state of supply of heat per unit time =λ. So according to the first condition. λt1=mc×100⇒mλ=t1100C...(i), and according to the second condition, λt2=mLf⇒mλ=t2Lf...(ii). By (i) and (ii) t1100C=t2Lf⇒101cal×100=55Lf⇒Lf=550calg−1
An experiment takes 10 min to raise the temperature of water in a container from 0^oC to 100^oC and another 55 min to convert it totally into steam…
JEE Main 2015 — Physics Thermodynamics
An experiment takes 10min to raise the temperature of water in a container from 0oC to 100oC 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(g∘C)−1 , the heat of vaporization according to this experiment will come out to be:
Held on 11 Apr 2015 · Verified 6 Jul 2026.
Options
- A
530calg−1
- B
550calg−1
- C
540calg−1
- D
560calg−1
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