CDS (II) 2023 General Knowledge · Question 37 of 120

Correct — D, T_F = (9/5)T_C + 32. Both scales are linear, so the conversion has the form of a straight line — a multiplier for the size of one degree and an added constant for where the scale starts. Fix the two constants from the two fixed points. Water freezes at 0 on the Celsius scale and at 32 on the Fahrenheit scale; it boils at 100 Celsius and 212 Fahrenheit. Between those points Celsius covers 100 degrees while Fahrenheit covers 212 minus 32, which is 180. So one Celsius degree equals 180/100 = 9/5 Fahrenheit degrees, which is the multiplier, and the offset is the freezing-point reading of 32, because that is the Fahrenheit value when the Celsius value is zero. Putting them together gives T_F = (9/5)T_C + 32. Test it at both ends: at T_C = 0 the expression returns 32, and at T_C = 100 it returns 180 plus 32, which is 212. Body temperature is a third check — 37 Celsius gives 66.6 plus 32, that is 98.6 Fahrenheit, the familiar figure.

• (a)T_F = (10/3)T_C + 32 — The offset is right but the slope is wrong. A multiplier of 10/3 would put the boiling point at 333 plus 32, about 365 Fahrenheit, instead of 212. The slope has to be the ratio of the two intervals, 180 to 100.
• (b)T_F = (5/9)T_C + 36 — Wrong on both counts. 5/9 is the inverted multiplier — it belongs to the reverse conversion, from Fahrenheit to Celsius — and 36 is not the Fahrenheit reading of the ice point. Testing it at 0 Celsius returns 36 rather than 32.
• (c)T_F = (9/5)T_C + 36 — The right slope with the wrong offset, and the most tempting of the three wrong options. At 0 Celsius it gives 36 Fahrenheit; water freezes at 32 Fahrenheit, so the whole line is shifted four degrees too high.

A temperature scale is defined by two fixed points and by how the interval between them is divided. The Celsius scale puts 0 at the ice point and 100 at the steam point at standard pressure, dividing the gap into 100 parts. The Fahrenheit scale puts the same two points at 32 and 212, dividing the same gap into 180 parts. Because both are linear in the same physical quantity, any reading on one converts to the other by a fixed multiplier plus a fixed offset. The Kelvin scale shares the Celsius degree size but starts at absolute zero, so T_K = T_C + 273.15.

The reliable method is never to memorise the formula but to rebuild it from the two fixed points, then test the result at a value you know. The slope must be 180 over 100, the ratio of the interval lengths, and the constant must be whatever Fahrenheit reads when Celsius reads zero. That immediately kills any option with 36 in it and any option whose multiplier is not 9/5. A related fact worth carrying is the single temperature at which the two scales read the same number: setting T_F equal to T_C gives T_C = (9/5)T_C + 32, which solves to minus 40. Both scales read minus 40 there, and examiners ask that version of the question regularly.

• Water freezes at 0 degrees Celsius and 32 degrees Fahrenheit; it boils at 100 degrees Celsius and 212 degrees Fahrenheit at standard pressure.
• The interval between the fixed points is 100 Celsius degrees but 180 Fahrenheit degrees, so one Celsius degree equals 9/5 Fahrenheit degrees.
• The conversion is T_F = (9/5)T_C + 32, and its inverse is T_C = (5/9)(T_F − 32).
• The two scales read the same number, minus 40, at one temperature only.
• The Kelvin scale uses the Celsius degree size but starts at absolute zero: T_K = T_C + 273.15.

• Using 5/9 in the Celsius-to-Fahrenheit direction; that multiplier belongs to the reverse conversion.
• Writing 36 instead of 32 for the offset — the offset is the Fahrenheit reading at the ice point.
• Forgetting to test the finished formula at a known value; one substitution of 100 Celsius eliminates every wrong option here.

Either as a straight recall of the conversion relation, or as a numerical — convert a given Fahrenheit reading to Kelvin, or find the temperature at which two scales read alike.

• practice — not a real PYQ

  At which temperature do the Celsius and Fahrenheit scales show the same numerical reading?

  • (a)0 degrees
  • (b)−40 degrees
  • (c)−273 degrees
  • (d)100 degrees

  Answer(b) −40 degrees — setting T_F equal to T_C in T_F = (9/5)T_C + 32 gives T_C = −40, the one point at which the two scales agree.

• practice — not a real PYQ

  A temperature of 45 degrees Celsius corresponds to which one of the following Fahrenheit readings?

  • (a)81 degrees F
  • (b)104 degrees F
  • (c)113 degrees F
  • (d)117 degrees F

  Answer(c) 113 degrees F — (9/5)(45) = 81, and 81 + 32 = 113.