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11th Standard English Medium Physics Subject Heat and Thermodynamics Book Back 1 Mark Questions with Solution Part - I updated Book back Questions

Published on: 13/05/2022

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Questions:

Multiple Choice Question

5 x 1 = 5

An ideal refrigerator has a freezer at temperature −12°C. The coefficient of performance of the engine is 5. The temperature of the air (to which the heat ejected) is

(a)

50°C

(b)

45.2°C

(c)

40.2°C

(d)

37.5°C

In an isochoric process, we have

(a)

W = 0

(b)

Q = 0

(c)

ΔU = 0

(d)

ΔT = 0

A distant star emits radiation with maximum intensity at 350 nm. The temperature of the star is

(a)

8280 K

(b)

5000 K

(c)

7260 K

(d)

9044 K

An ideal gas is taken from (P_i, V_i) to (P_f, V_f) in three different ways. Identify the process in which the work done on the gas the most.

(a)

Process A

(b)

Process B

(c)

Process C

(d)

Equal work is done in Process A, B & C

When you exercise in the morning, by considering your body as thermodynamic system, which of the following is true?

(a)

ΔU > 0, W > 0,

(b)

ΔU < 0, W > 0,

(c)

ΔU < 0, W < 0,

(d)

ΔU = 0, W > 0,

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Answers Key:

Multiple Choice Question

5 x 1 = 5

\(\mathrm{COP}=\frac{T_{L}}{T_{H}-T_{L}}\)
\(\mathrm{T}_{\mathrm{L}}=-12+273 =261 \mathrm{~K} \)
\(5=\frac{261}{T_{H}-261} \)
\(\therefore 5\left(T_{H}-261\right) =261 \)
\(5 \mathrm{~T}_{\mathrm{H}}-1305 =261 \)
\(5 \mathrm{~T}_{\mathrm{H}} =261+1305 =1566 \)
\(\therefore T_{H} =\frac{1566}{5} \)
\(=313.2 \mathrm{~K} \)
\(\mathrm{~T}_{\mathrm{H}}=313.2-273 =40.2^{\circ} \mathrm{C} \)

Work done in an isochoric process is zero.

\(\lambda_{m} T =\mathrm{b} \)
\(\therefore T =\frac{2.898 \times 10^{-3}}{350 \times 10^{-9}}\)
\(\mathrm{~T} =0.00828 \times 10^{6} \)
\(=8280 \mathrm{~K} \)

When work is done on the gas, compression takes place. P_fwill be increased and remains constant.
Therefore Process B

\(\Delta Q=\Delta U+\Delta W\)
When, exercise is performed internal energy is utilised (\(\Delta\)U < 0) so it will decrease. But as intemal energy is utilised, exercise ( \(\Delta\)W ) will be more efficient.
\(\therefore \Delta W>0 \ \therefore \Delta U <0 \mathrm{~W}>0\)