alcoln Andson PHY 161 UNIVERSITY PHYSICS TEST I January 2018 MULTIPLE CHOICE: Ma

Question

alcoln Andson PHY 161 UNIVERSITY PHYSICS TEST I January 2018 MULTIPLE CHOICE: Mark the hest choice on the answer sheet provided. 1. Heat is a. the amount of thermal energy in an object. b. an invisible fluid-like substance that flows. c. the energy that moves from a hotter object to a colder object due to a temperature difference 2. Heat has the same units as a. work b. temperature c. energy/time d. energy/volume 3·The heat of fusion of water is 333 kJ/kg. This means that 333 kJ of heat are required To a. melt I kg of ice at its melting point. b. raise the temperature of 1 kg of water by 1 K. c turn I kg of water to steam at its boiling point. d. increase the internal energy of 1 kg of water by 1 kJ. 4. Five kilograms of water at 100°C is converted into steam. The heat added to the water is (Lf 333 kJ/kg, Lv-2256 kJ/kg) a. 1,665 kJ b. 11.3 X 103 kJ c. 33.3 kJ d. 1.13 X 106k 5. How much heat must be removed from 2 kg of water at 20°C to convert it into ice at 0°C? (cw -4190 J/(kg-K, Le (water) 333,000 J/kg). a. 167.6 kJ b. 666 kJ c. 834 kJ d. 2000 kJ 6. When heat is added to boiling water, its temperature a. increases b. decreases c. stays the same 7. The specific heat of lead is 128 J/(kg-K). If0.3 kg of lead at 100°C is mixed with 0.1 kg of water (o 4190 J/(kg-K) at 70°C in an insulated container, the final temperature of the mixture is a. 100 b. 85.5 c. 79.5 °C d. 72.5 8. An ideal gas is maintained at a constant pressure of 70,000 Pa during an isobaric process and its volume increases by 0.2 m2. How much work is done by the gas? a. -14,0000 J b. 14,000J c. 35,000J d. -35,000.J 9, Five moles of an ideal gas expands isothermally at 97°C to three times its initial volume. The work done by the gas is (R-8.31 J/mole-K) d. 1.0 X 10J

Explanation / Answer

1)

(a) The energy that moves from a hotter object to a colder object due to a temperature difference.

2)

(a) work

3)

(a)

4)

(b)

Q = mLv = 5 x 2256 = 11.3 x 10^3 kJ

5)

(c)

Q = mcdT + mLf

= 2[(20)(4190) + 333000]

= 833,600J

6)

(c)

7) (d)

Let T be the final temp.

Heat lost by lead = heat gained by water

=> 0.3(128)(100-T) = 0.1(4190)(T-70)

=> 3840 - 38.4T = 419 T - 29330

=> T = 72.5

8) (b)

W = PdV

= 70000(0.2)

= 14000J

9) (a)

W = nRT ln(V2/V1)

= (5)(8.31)(97+273 K) ln(3)

= 1.7 x 10^4 J

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