Thermometers
Q1

Convert 93 °C into
(i) Kelvin
(ii) Fahrenheit scales

Sol:
(i) Temperature in Kelvin = 273 + temp in (°C)
= 273 + 93
= 366 K.
(ii) F – 32/9 = C/5
∴ F – 32 = 9/5 ×C
= 9/5 × 93
= 167.4
F = 167.4 + 32
= 199.4 °F
Q2

Convert 102.4° F into
(i) Celsius
(ii) Kelvin scale

Sol:
(i) F – 32/9 = C/5
∴ C = 5/9 (F – 32)
= 5/9 (102.4 – 32)
= 5/9 × 70.4
= 39.1° C
(ii) Temperature in Kelvin = 273 + temp in (°C)
= 273 + 39.1
= 312.1 K.
Q3

Convert 183 K into
(i) Celsius
(ii) Fahrenheit scale

Sol:
(i) Temperature in Kelvin = 273 + temp in (°C)
183 = 273 + temp in (°C)
∴ Temp. on Celsius scale = 183 – 273 .
= – 90° C
(ii) F – 32/9 = C/5
∴ F – 32 = 9c/5
= 9/5 × – 90
= – 162
F = – 162 + 32
= – 130° F
Q4

The mercury thread rises by 5/l6 parts between the two standard points on Celsius scale, when thermometer is placed in warm milk. Calculate the temperature in
(i)° C
(ii) K
(iii) Fahrenheit scale

Sol:
(i) 5/16 of Celsius scale = 5/16 ×100 ° C
= 31.25° C
(ii) Temp on Kelvin scale = 31.25 + 273
= 304.25 K
(iii) Temp on Fahrenheit scale is F – 32/9 = C/5
∴ F – 32 = 9/5 ×C
F = 9/5 × C + 32
= 9/5 × 31.25 + 32
= 88.25° F
Q5

The mercury thread falls by 15/16 parts between two standard points on Fahrenheit scale, when boiling water at 82 cm pressure is cooled to room temperature. Calculate the room temperature in
(i) Fahrenheit
(ii) Celsius
(iii) Kelvin scales

Sol:
No. of degrees between standard points on Fahrenheit scale = (212 – 32)
= 180° F
(i) 15/16 parts of Fahrenheit scale = 15/16 × 180 ° C
= 168.75° F
Room temp. on Fahrenheit = (180 – 168.75) + 32
= 43.25° F
(ii) F – 32/9 = C/5
∴ C = 5/9 (F – 32)
= 5/9 (43.25 – 32)
= 6.25° C
(iii) Temperature in K = C + 273
= 6.25 + 273
= 279.25 K.
Q6

A thermometer has its fixed points marked 93 and – 5. What is the correct temperature in °C, when this thermometer reads 55°?

Sol:
Interval between fixed points on faulty thermometer = [93 – (– 5)]
= 98°
Let x be the correct temp. in Celsius.
Temp. on faulty thermometer = 55° C.
∴ Reading on faulty thermometer – starting point / Interval between fixed points = x/100
55 – ( – 5)/98 = x/100
60/98 = x/100
x = 6000/98
= 61.23° C.
Q7

When will the numerical values of Fahrenheit thermometer be equal and opposite to values on Celsius scale?

Sol:
By the condition of question, x° F = – x° C
Applying F – 32/9 = C/5
x – 32/9 = – x/5
∴ 5x – 160 = – 9x
14x = 160
∴ x = 160/14
= 11.42° F

∴ 11.42° F (or) – 11.42° C are numerically equal and opposite.