If the substance is heated at a constant rate d t d Q

At constant power, time is proportional to heat. From the liquid segment: 130 − 106 Q − 106 = 20 − ( − 10 ) 10 − ( − 10 ) Q = 122 kcal From the gas segment: 346 − 330 Q ′ − 330 = 60 − 20 40 − 20 Q ′ = 338 kcal The two intervals require: Δ Q 1 = 122 kcal Δ Q 2 = 338 − 122 = 216 kcal Δ Q 2 > Δ Q 1 The second interval therefore takes longer.

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Passage / Comprehension

The figure below shows a plot of temperature vs. heat energy supplied to $1.00 kg$ of some substance. The substance has melting point of $- 10. 0^{\circ} C$ and is a gas at temperatures above $20. 0^{\circ} C$ .

If the substance is heated at a constant rate $\frac{d Q}{d t}$

Options:

Answer:

Solution:

At constant power, time is proportional to heat. From the liquid segment:
$\frac{Q - 106}{130 - 106} = \frac{10 - ( - 10 )}{20 - ( - 10 )}$
$Q = 122 kcal$
From the gas segment:
$\frac{Q ^{'} - 330}{346 - 330} = \frac{40 - 20}{60 - 20}$
$Q^{'} = 338 kcal$
The two intervals require:
$Δ Q_{1} = 122 kcal$
$Δ Q_{2} = 338 - 122 = 216 kcal$
$Δ Q_{2} > Δ Q_{1}$
The second interval therefore takes longer.

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:Calorimetry

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