Thermal Properties of Matter

JEE Main level, numerical answer, medium

66 of 126

Question 66 of 126

A steam engine intakes 50g50g of steam at 100∘C100^{\circ}C per minute and cools it down to 20∘C20^{\circ}C. If latent heat of vaporization of steam is 540calg−1540cal{g}^{-1}, then the heat rejected by the steam engine per minute is _____ ×103cal\times {10}^{3}cal
(Given : specific heat capacity of water : 1calg−1C∘−11cal{g}^{-1}{C^{\circ}}^{-1})

Show answer and solution

Answer: 31

Heat rejected by the steam engine per minute will be equal to latent heat per minute plus heat rejected by the water to cool down.
∆H∆t=∆m∆tL+∆m∆ts∆T⇒∆H∆t=50×540+50×1×80=31000calmin⁡−1=31×103calmin⁡−1\dfrac{∆H}{∆t}=\dfrac{∆m}{∆t}L+\dfrac{∆m}{∆t}s∆T \Rightarrow \dfrac{∆H}{∆t}=50\times 540+50\times 1\times 80=31000cal{\min }^{-1}=31\times {10}^{3}cal{\min }^{-1}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.

More Physics practice

JEE Main Physics practice question on Thermal Properties of Matter (Calorimetry), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.