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Chapter 15: 8MCQ (page 412)

On a very hot day, could you cool your kitchen by leaving the refrigerator door open?

(a) Yes, but it would be very expensive.

(b) Yes, but only if the humidity is below 50%.

(c) No, the refrigerator would exhaust the same amount of heat into the room as it takes out of the room.

(d) No, the heat exhausted by the refrigerator into the room is more than the heat the refrigerator takes out of the room.

Short Answer

Expert verified

The correct option is (d).

Step by step solution

01

Principle of refrigerator

A refrigerator is a device to cool a lower temperature region by extracting heat from that region.

By doing work, heat is taken out from a low-temperature region (such as inside a refrigerator), and a greater amount of heat is exhausted to a high-temperature region (the room).

02

Explanation

On a very hot day, if you open your refrigerator door, it will not be possible to cool the whole kitchen because the heat exhausted by the refrigerator into the room is more than the heat the refrigerator takes out of the room.

Hence, option (d) is correct.

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Most popular questions from this chapter

Question: (a) At a steam power plant, steam engines work in pairs, the heat output of the first one being the approximate heat input of the second. The operating temperatures of the first are 750°C and 440°C, and of the second 415°C and 270°C. If the heat of combustion of coal is \({\bf{2}}{\bf{.8 \times 1}}{{\bf{0}}^{\bf{7}}}\;{{\bf{J}} \mathord{\left/{\vphantom {{\bf{J}} {{\bf{kg}}}}} \right.} {{\bf{kg}}}}\) at what rate must coal be burned if the plant is to put out 950 MW of power? Assume the efficiency of the engines is 65% of the ideal (Carnot) efficiency. (b) Water is used to cool the power plant. If the water temperature is allowed to increase by no more than 4.5 C°, estimate how much water must pass through the plant per hour.

Question:(II) A certain power plant puts out 580 MW of electric power. Estimate the heat discharged per second, assuming that the plant has an efficiency of 32%.

A “Carnot” refrigerator (the reverse of a Carnot engine) absorbs heat from the freezer compartment at a temperature of -17°C and exhausts it into the room at 25°C.

(a) How much work would the refrigerator do to change 0.65 kg of water at 25°C into ice at -17°C.

(b) If the compressor output is 105 W and runs 25% of the time, how long will this take?

Question: (I) The low temperature of a freezer cooling coil is\({\bf{ - 8^\circ C}}\)and the discharge temperature is\({\bf{33^\circ C}}\). What is the maximum theoretical coefficient of performance?

Question: (III) The PV diagram in Fig. 15–23 shows two possible states of a system containing 1.75 moles of a monatomic ideal gas. \(\left( {{P_1} = {P_2} = {\bf{425}}\;{{\bf{N}} \mathord{\left/{\vphantom {{\bf{N}} {{{\bf{m}}^{\bf{2}}}}}} \right.} {{{\bf{m}}^{\bf{2}}}}},\;{V_1} = {\bf{2}}{\bf{.00}}\;{{\bf{m}}^{\bf{3}}},\;{V_2} = {\bf{8}}{\bf{.00}}\;{{\bf{m}}^{\bf{3}}}.} \right)\) (a) Draw the process which depicts an isobaric expansion from state 1 to state 2, and label this process A. (b) Find the work done by the gas and the change in internal energy of the gas in process A. (c) Draw the two-step process which depicts an isothermal expansion from state 1 to the volume \({V_2}\), followed by an isovolumetric increase in temperature to state 2, and label this process B. (d) Find the change in internal energy of the gas for the two-step process B.
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