Calculate the temperature which has same numeral value on celsius and Fahrenheit scale.

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Q. Calculate the temperature which has same numeral value on celsius and Fahrenheit scale.

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Concept understanding — Temperature Scale Conversion

Temperature Scale Conversion

You already know temperature as a measure of hotness or coldness. But here's the thing: different countries and different scientific fields measure that same hotness using different numbers. Water boils at 100 on the Celsius scale, but at 212 on the Fahrenheit scale. That's not because the water is different — it's because the rulers are different.

The core idea

A temperature scale is just a number line someone decided to draw on the same physical reality. Converting between scales means finding the number on one number line that corresponds to the same physical hotness as a given number on another number line.

Think of it like this: if you measure a table's length in feet and get 6, and I measure it in metres and get 1.83, we're both describing the same table. Temperature conversion works exactly the same way — same physical state, different numbers.

The three main scales you'll meet

Scale	Freezing point of water	Boiling point of water	Used by
Celsius (°C)	0	100	Most of the world, science
Fahrenheit (°F)	32	212	USA, a few other countries
Kelvin (K)	273.15	373.15	All scientific work

Notice something: Celsius and Kelvin have the same step size — a change of 1°C is exactly a change of 1 K. They just start at different places. Fahrenheit has a smaller step size (180 steps between freezing and boiling, instead of 100).

The conversion formulas

°F = \frac{9}{5}(°C) + 32

These aren't magic. Each one comes from the simple idea of matching two number lines.

Why the formulas look like that

Between freezing and boiling water:

Celsius has 100 steps (0 to 100)
Fahrenheit has 180 steps (32 to 212)

So one Celsius step is $\frac{180}{100} = \frac{9}{5}$ Fahrenheit steps. That's the $\frac{9}{5}$ factor. The $+32$ just shifts the starting point — because 0°C doesn't correspond to 0°F, it corresponds to 32°F.

For Kelvin: since the step size is identical to Celsius, you just add the offset. The 273.15 comes from the fact that absolute zero (the coldest possible temperature) is 0 K, which is $-273.15$ °C.

A worked example

Convert 25°C to Fahrenheit.

Step 1: Multiply by $\frac{9}{5}$ .

25 \times \frac{9}{5} = 25 \times 1.8 = 45

Step 2: Add 32.

45 + 32 = 77

So 25°C = 77°F. A warm spring day in Celsius terms is 77°F — same weather, different number.

Tip

For quick mental estimates: double the Celsius and add 30. It's not exact (you get 80 instead of 77 here), but it's close enough for everyday "should I wear a jacket?" decisions.

The one thing students mess up …

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We find the temperature where the numerical value is identical on both scales by setting $C = F$ in the conversion formula. The answer is −40°.

The Celsius and Fahrenheit scales measure the same physical quantity—temperature—but use different reference points and step sizes. The Celsius scale sets water's freezing point at 0° and boiling at 100°, while Fahrenheit uses 32° and 212° for the same landmarks. This means the scales diverge everywhere except at one special crossing point.

The conversion formula connecting them is:

F = \frac{9}{5}C + 32

or equivalently,

C = \frac{5}{9}(F - 32)

We want the temperature where the number is the same on both thermometers—that is, where $C = F$ numerically. Let's call this common value $x$ .

Step-by-step solution:

Set up the equation. If the numerical value is the same, then $C = F = x$ . Substitute into the conversion formula:

x = \frac{9}{5}x + 32

Isolate $x$ . Subtract $\frac{9}{5}x$ from both sides:

x - \frac{9}{5}x = 32

Combine the terms on the left. Write $x$ as $\frac{5}{5}x$ :

\frac{5}{5}x - \frac{9}{5}x = 32

-\frac{4}{5}x = 32

Solve for $x$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
$^\circ$
›Reveal solutionSolution
The key idea is to set up the relationship between Fahrenheit and Celsius using the standard conversion formula, then impose the condition that the Fahrenheit reading is 90% more than the Celsius reading. Solving yields 608 °F, so the correct option is (C).

Concept and Intuition

The Fahrenheit and Celsius scales are linearly related: $F = \frac{9}{5}C + 32$ . The phrase “90% more than” means the Fahrenheit value is 190% of the Celsius value (i.e., $F = 1.9C$ ). We equate these two expressions for $F$ and solve for $C$ , then find $F$ . The trick is to avoid misinterpreting “90% more” as simply $F = C + 0.9C$ — which is exactly what we do, but we must be careful with the constant 32.

Step-by-step solution

Write the standard conversion The relationship between Fahrenheit ( $F$ ) and Celsius ( $C$ ) is:

$F = \frac{9}{5}C + 32.$

Translate the condition “Fahrenheit reading is 90% more than the Celsius reading” means:

$F = C + 0.9C = 1.9C.$
(If something is 90% more, you add 90% of itself, so it becomes 190% of the original.)

Set the two expressions equal

$1.9C = \frac{9}{5}C + 32.$
Note that $\frac{9}{5} = 1.8$ . So:
$1.9C = 1.8C + 32.$

Solve for $C$ Subtract $1.8C$ from both sides:

$0.1C = 32 \quad \Rightarrow \quad C = 320.$
So the Celsius temperature is $320^\circ\text{C}$ .

Find the corresponding Fahrenheit …
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TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
$^\circ$
›Reveal solutionSolution
The key idea is to set up the relation $F = C + 0.9C = 1.9C$ and then use the standard conversion formula $F = \frac{9}{5}C + 32$ to solve for $C$ , then find $F$ . The result is $F = 608^\circ\text{F}$ , so option (A) is correct.

Concept and intuition:

The problem asks for a temperature where the Fahrenheit reading is 90% more than the Celsius reading. That means Fahrenheit is not just 90% of Celsius, but Celsius plus 90% of Celsius, i.e., $F = C + 0.9C = 1.9C$ . We also know the standard linear relationship between the two scales: $F = \frac{9}{5}C + 32$ . Equating these two expressions for $F$ lets us solve for the Celsius temperature, and then find the corresponding Fahrenheit.

Step-by-step solution:

Translate the condition into an equation. "90% more than the reading on Celsius scale" means:

$F = C + 0.9C = 1.9C$
This is the first relation.

Write the standard conversion formula. The relationship between Fahrenheit and Celsius is:

$F = \frac{9}{5}C + 32$
This is the second relation.

Set the two expressions for $F$ equal. Since both represent the same Fahrenheit temperature, we have:

$1.9C = \frac{9}{5}C + 32$

Solve for $C$ . Convert $\frac{9}{5}$ to decimal: $\frac{9}{5} = 1.8$ . Then:

$1.9C = 1.8C + 32$
Subtract $1.8C$ from both sides:
$0.1C = 32$
Divide by 0.1:

$C = 320$ …
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TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
$39^\circ \mathrm{C}$
›Reveal solutionSolution
The coefficient of performance (COP) of a refrigerator relates the heat extracted from the cold reservoir to the work input. For an ideal (Carnot) refrigerator, COP = $T_c/(T_h - T_c)$ . Using COP = 5 and $T_h = 39^\circ \mathrm{C} = 312\,\mathrm{K}$ , we solve for $T_c$ and convert to Celsius, getting $-13^\circ \mathrm{C}$ .

The key idea is that the coefficient of performance of a refrigerator is defined as the ratio of the heat removed from the cold space to the work supplied. For an ideal refrigerator operating on a Carnot cycle, this COP depends only on the absolute temperatures of the hot and cold reservoirs. The problem gives the COP and the room temperature (hot reservoir), so we can find the inside temperature (cold reservoir) — but only if we assume the refrigerator is ideal. In practice, real refrigerators have lower COP, but exam problems always treat the given COP as the Carnot COP unless stated otherwise.

Let’s work through it step by step.

Convert temperatures to Kelvin.

The room temperature is $39^\circ \mathrm{C}$ .

$T_h = 39 + 273 = 312\,\mathrm{K}$ .

The inside temperature $T_c$ is unknown, in Kelvin.

Recall the formula for COP of a Carnot refrigerator.

For a refrigerator,

$\text{COP} = \frac{Q_c}{W} = \frac{T_c}{T_h - T_c}$
where $T_c$ and $T_h$ are absolute temperatures. This formula comes from the fact that for a Carnot cycle, $Q_c/T_c = Q_h/T_h$ and $W = Q_h - Q_c$ , leading directly to the expression above.

Plug in the given values. We have COP = 5 and $T_h = 312\,\mathrm{K}$ . So

$5 = \frac{T_c}{312 - T_c}$

Solve for $T_c$ . Multiply both sides by $(312 - T_c)$ :

$5(312 - T_c) = T_c$ $1560 - 5T_c = T_c$ $1560 = 6T_c$ $T_c = \frac{1560}{6} = 260\,\mathrm{K}$

Convert back to Celsius. $T_c = 260 - 273 = -13^\circ \mathrm{C}$ . …
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TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
$160\ ^\circ \mathrm{F}$
›Reveal solutionSolution
The key idea is to set up the relationship between Fahrenheit and Celsius using the conversion formula, then impose the condition that the Fahrenheit reading is twice the Celsius reading. Solving gives $320^\circ\text{F}$ , so the correct option is (C).

The problem asks for a Fahrenheit temperature that is exactly twice the corresponding Celsius temperature. This is a classic linear relationship problem: the two scales are linked by the formula $F = \frac{9}{5}C + 32$ . The trick is not to guess but to translate the English condition into algebra.

Set up the condition We want the Fahrenheit reading to be twice the Celsius reading:

$F = 2C$

Use the conversion formula The standard conversion from Celsius to Fahrenheit is:

$F = \frac{9}{5}C + 32$

Substitute and solve Replace $F$ in the conversion formula with $2C$ :

$2C = \frac{9}{5}C + 32$
Subtract $\frac{9}{5}C$ from both sides:
$2C - \frac{9}{5}C = 32$
Write $2C$ as $\frac{10}{5}C$ :
$\frac{10}{5}C - \frac{9}{5}C = 32 \quad \Rightarrow \quad \frac{1}{5}C = 32$
Multiply both sides by 5:

$C = 160$ …
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TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
$T_1 = 400\,\text{K}$
›Reveal solutionSolution

The engine's work drives the refrigerator. Relating the heat flows through the shared work with the Carnot relation $Q_{\text{hot}}/Q_{\text{cold}} = T_{\text{hot}}/T_{\text{cold}}$ gives $\frac{Q_3}{Q_1} = 1.75$ , so the correct option is (D).

An ideal Carnot engine operates between $T_1 = 400\,\text{K}$ and $T_2 = 200\,\text{K}$ ; its work output drives a Carnot refrigerator operating between $T_3 = 350\,\text{K}$ (hot) and $T_4 = 250\,\text{K}$ (cold). Here $Q_1$ is the heat the engine absorbs from its hot reservoir, and $Q_3$ is the heat the refrigerator rejects at $T_3$ .

1. Engine. For a Carnot engine, $\frac{Q_2}{Q_1} = \frac{T_2}{T_1} = \frac{200}{400} = 0.5$ , so the work is

$W = Q_1 - Q_2 = 0.5\,Q_1.$ …
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Q.Calculate the temperature which has same numeral value on celsius and Fahrenheit scale.

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