Newton's Law of Cooling NEET 2014: Why the Answer Is 45 °C

What is the answer to the Newton’s law of cooling NEET 2014 question?

The surroundings temperature is 45 °C, option A. The Newton's law of cooling NEET 2014 question is solved by comparing temperature excess above the surroundings, not the water’s thermometer readings.

A sample of water starts at 70 °C, reaches 60 °C after five minutes, and reaches 54 °C after another five minutes. Find the surroundings temperature.

This Physics question belongs to Properties of Solids and Liquids. The question bank tags it hard, with an expected solving time of 90 seconds, not an official exam limit.

Which temperatures belong in the equal-interval ratio?

Use each water temperature minus the surroundings temperature. Under unchanged Newtonian cooling conditions, equal time intervals retain the same fraction of temperature excess. They do not remove the same number of degrees each time.

Let theta represent the constant surroundings temperature in °C. The three excess temperatures, also in °C, are:

For each interval, divide the excess at its end by the excess at its start. The official solution therefore begins with:

Both intervals last five minutes, so their retained fractions are equal. The cooling constant and interval length determine that fraction, but neither needs to be calculated here.

Keep the order consistent: end excess over start excess on both sides. Using raw Celsius readings would ignore the surroundings, which set the reference temperature for cooling.

How does the equation give 45 °C?

Cross-multiplication gives an apparent quadratic, but the squared terms cancel, leaving a linear equation. There is no need for a quadratic formula.

Start by cross-multiplying:

Expand the left side:

Expand the right side:

Equate the expanded expressions:

Cancel the identical squared term from both sides: $3600-120\theta =3780-124\theta$

Collect the unknown terms on the left and constants on the right: $124\theta -120\theta =3780-3600$

Simplify: $4\theta =180$

Therefore: $\theta =45{}^{\circ }C$

Correct answer: A) 45 °C. Keep the cross-products visible while expanding to avoid a sign error.

How can you check 45 °C in both intervals?

Substituting 45 °C gives excess temperatures of 25 °C, 15 °C and 9 °C. Both five-minute intervals retain exactly three-fifths of the starting excess, so the answer satisfies the cooling condition independently of the algebra.

The absolute temperature falls are unequal: 10 °C first, then 6 °C. Cooling slows as the water approaches the surroundings temperature because the temperature excess becomes smaller.

Also, 45 °C lies below all three recorded water temperatures, as expected for this cooling process. That physical check is necessary, but it cannot select the answer alone: every listed option lies below 54 °C.

How can a wrong ratio produce option C, 42 °C?

Option C can result from equating a valid ratio of temperature falls to an invalid ratio of Celsius readings. The successive falls are: $70-60=10{}^{\circ }C$ $60-54=6{}^{\circ }C$

The incorrect setup is:

This wrongly identifies surroundings temperature divided by initial Celsius temperature with the fraction retained between intervals. The ratio of successive falls does equal the excess-retention factor in this equal-interval model:

That does not justify using surroundings temperature divided by initial temperature. The fraction connects successive excesses above the surroundings:

Substitution rejects 42 °C directly. The retained fractions are:

The fractions are unequal, contradicting the supplied cooling model. This method error produces a listed distractor, not a valid alternative solution.

How do you apply this method to two related cooling questions?

Keep using temperature excess, but distinguish an interval’s retained fraction from an instantaneous cooling rate. These are original follow-up practice questions based on the supplied setup, not additional verified PYQs.

What is the water temperature five minutes after it reaches 54 °C?

The expected temperature is 50.4 °C, assuming the same Newtonian cooling conditions continue. At 54 °C, the water is still 9 °C above its 45 °C surroundings, and another five-minute interval retains three-fifths of that excess.

The new excess is:

Add the surroundings temperature to convert excess back into the water temperature:

Subtracting the previous 6 °C fall again would incorrectly assume a constant cooling rate. Under the same model, the expected next fall is smaller:

What is the ratio of instantaneous cooling-rate magnitudes at 70 °C and 54 °C?

The required ratio is 25 to 9. Newton’s law makes the instantaneous cooling-rate magnitude proportional to temperature excess above the surroundings.

The cooling constant is unchanged, so it cancels in the ratio:

These are rates at two particular temperatures, not average rates over five-minute intervals. The original interval averages are instead:

Before forming a rate ratio, mark whether the question asks for an instantaneous rate or an interval average. Use temperature excesses for the former and temperature fall divided by elapsed time for the latter.

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