# Chemical Engineering Questions and Answers – Temperature

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This set of Basic Chemical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Temperature”.

1. Which of the following is not a unit of temperature?
a) Celsius
b) Fahrenheit
c) Therm
d) Rankine

Explanation: Them is a unit of energy, 1 therm = 105 BTU.

2. Which of the following is greatest?
a) 10oC
b) 10oR
c) 10oF
d) 10 K

Explanation: 10oC = 283 K = 50oF = 510oR, => 10oC is greatest.

3. Absolute zero temperature is approximately equal to?
a) -160oR
b) -273oR
c) -373oR
d) -460oR

Explanation: Absolute zero temperature = 0 K = -273oC = -460oR.
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4. If change in temperate is 212 degree Fahrenheit, what is the change in temperature in degree Rankine?
a) 100oR
b) 212oR
c) 273oR
d) 460oR

Explanation: Since ∆TR = ∆TF, => 212oF = 212oR.

5. If change in temperate is 180 degree Rankine, what is the change in temperature in degree Celsius?
a) 100oC
b) 180oC
c) 212oC
d) 273oC

Explanation: ∆TR = 1.8∆TC, => Change in temperature = 180/1.8 = 100oC.

6. Let us say temperature of a body in degree Celsius and degree Rankine are related as TC = 50 + 3TR, what is the change in temperature in Kelvin, if the change in temperature in degree Rankine is 100oR?
a) 300 K
b) 180 K
c) 100 K
d) 90 K

Explanation: ∆TC = 3∆TR, => ∆TC = 3*100 = 300oC, => ∆TK = ∆TC = 300 K.

7. What is the value of 100oC in degree Fahrenheit?
a) 212oF
b) 460oF
c) 672oF
d) 884oF

Explanation: TF = (9/5)100 + 32 = 212oF.

8. What is the value of 273 K in degree Rankine?
a) 0oR
b) 32oR
c) 492oR
d) 672oR

Explanation: TR = (9/5)(273 – 273) + 492 = 492oR.

9. Volume of an object is dependent upon its temperature in degree Celsius as V = a + bTC, what is the dependence of volume upon temperature in degree Rankine?
a) V = a + 32b + 9bTR/5
b) V = a + 492b + 9bTR/5
c) V = a + 492b + 5bTR/9
d) None of the mentioned

Explanation: TC = 492 + 9TR/5, => V = a + b(492 + 9TR/5), => V = a + 492b + 9bTR/5.

10. Mass and volume of an object is dependent on its temperature in kelvin as m = 5t2 kg and v = (400 + 600t) m3 respectively, then what is the density of the object at t = 27oC?
a) 5 kg/m3
b) 10 kg/m3
c) 15 kg/m3
d) 20 kg/m3

Explanation: Density of object = dm/dv = (dm/dt)/(dv/dt) = 10t/600 = t/60, => density of object at t = 27oC = 300 K is D = 300/60 = 5 kg/m3.

11. Heat capacity of a substance is given by C = 800 – 1600T + 25T2, where T is temperature in degree Fahrenheit, what is the heat capacity if temperature is in degree Celsius?
a) 81T2 + 24800
b) 81T2 – 24800
c) 25T2 + 24800
d) 25T2 – 24800

Explanation: Temperature in degree Fahrenheit, TF = (9/5)TC + 32, replacing TF in C, => C = 81T2 – 24800.

12. Heat capacity of a substance is given by C = BT2, where T is in degree Fahrenheit and the value of B is 50, what is the value of B if T is in degree Celsius?
a) 81
b) 25
c) 162
d) 100

Explanation: TF = 9TF/5 + 32 => coefficient of T2 in C will be B*(9/5)2 = 50*81/25 = 162.

13. What is the temperature 100oC in degree Rankine?
a) 100oR
b) 212oR
c) 560oC
d) 672oC

Explanation: Temperature = 9(100)/5 + 32 + 460 = 672oR.

14. What is the 100oC in degree Fahrenheit?
a) 0oF
b) 100oF
c) 212oF
d) 460oF

Explanation: Temperature = 9(100)/5 + 32 = 212oF.

15. What is the 40oF in degree Rankine?
a) 313oR
b) 460oR
c) 500oR
d) 672oR

Explanation: Temperature = 40 + 460 = 500oR.

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