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NEWTONS LAW OF COOLING - Quiz
Contributed by: TAIWO
  • 1. Newton’s law of cooling states that the rate of loss of heat of a body is directly proportional to
A) its volume
B) its specific heat capacity
C) The temperature difference between the body and the surroundings
D) its mass
  • 2. In Newton’s law of cooling, T s represents:
A) steam temperature
B) surrounding temperature
C) surface area
D) standard temperature
  • 3. Newton’s law of cooling is most accurate for
A) small temperature differences
B) Absolute zero
C) large temperature differences
D) Melting processes
  • 4. A body cools from 90°C to 70°C while the surrounding temperature is 30°C. The average excess temperature is
A) 20°C
B) 50°C
C) 60°C
D) 40°C
  • 5. Which factor increases the rate of cooling?
A) lower emissivity
B) Larger temperature difference
C) smaller surface area
D) vacuum surroundings
  • 6. A correction in calorimetry is needed because
A) Heat is lost to the surrounding
B) Heat is gained from the surroundings
C) Water changes mass
D) Thermometers are inaccurate
  • 7. A liquid cools faster when
A) Air movement increases
B) Surface area decreases
C) Air movement increases
D) Mass increases
  • 8. A hot body loses 300 J in 2 minutes. Find heat loss per second
A) 150 J/s
B) 2.5 J/s
C) 600 J/s
D) 1.5 J/s
  • 9. A body cools from 90°C to 85°C in 1 minute. What is the temperature drop
A) 2°C
B) 10°C
C) 15°C
D) 5°C
  • 10. A liquid at 70°C is in a room at 30°CIf k=0.05 per seconds, calculate the rate of cooling.
A) 2°C/s
B) 1°C/s
C) 4°C/s
D) 3°C/s
  • 11. The purpose of cooling correction is to obtain the:
A) Mass of the calorimeter
B) Actual quantity of heat exchanged
C) Lowest temperature
D) Room temperature
  • 12. Cooling correction is applied in calorimetry because
A) Heat is lost to the surroundings during the experiment
B) The calorimeter changes colour
C) The thermometer expands
D) Water evaporates completely
  • 13. In calorimetry, cooling correction is usually
A) Multiplied by mass
B) Subtracted from the observed temperature rise
C) Subtracted from the observed temperature rise
D) Added to the observed temperature rise
  • 14. A calorimeter loses heat mainly through
A) Magnetism
B) Radiation and convection
C) Compression
D) Reflection
  • 15. A temperature rise recorded is 15°C. If the cooling correction is +1°C, calculate the percentage correction.
A) 6.7%
B) 10%
C) 15%
D) 5%
  • 16. A temperature rise recorded is 15°C. If the cooling correction is +1°C, calculate the percentage correction.
A) Volume
B) Pressure
C) Quantity of Heat accurately
D) Density
  • 17. During a calorimetry experiment, the observed temperature rise is smaller than the true value because:
A) The calorimeter melts
B) Heat escapes to the surrounding
C) Temperature becomes constant immediately
D) Water gains mass
  • 18. The unit of quantity of heat is
A) Pascal
B) Watt
C) Newton
D) Joule
  • 19. An observed final temperature is 48.2°C and cooling correction is +0.5°C. Find the corrected final temperature.
A) 49.2°C
B) 47.7°C
C) 48.0°C
D) 48.7°C
  • 20. A body cools from 60°C to 50°C in 4 minutes. Room temperature is 20°C. Find average excess temperature.
A) 30°C
B) 35°C
C) 25°C
D) 45°C
  • 21. In the formula, what does the variable T(s) represent?
A) The time taken to reach thermal equilibrium
B) The starting temperature of the object
C) The constant temperature of the surrounding environment (surroundings)
D) The specific heat capacity of the material
  • 22. What does the variable T(t) represent in the equation
A) The time it takes for the temperature to drop to 0 degrees Celsius
B) The total heat energy lost by the system
C) The temperature of the object at any specific time t
D) The melting point of the polymer sample
  • 23. If a polymer melt at 180 degree Celsius is placed in a room at 25 degree Celcius. what is the value of (T(o) - T(s)?
A) 155 degree Celcius
B) 205 degree Celcius
C) 180 degree Celcius
D) 25 degree Celcius
  • 24. If T(s)= 25 degree Celcius, T(o) = 125 degree Celcius, and k= 0.1 per min, what is the temperature after 10 mins? (Take e-1= 0.368)
A) 61.8 degree Celcius
B) 25.0 degree Celcius
C) 75.2 degree Celcius
D) 36.8 degree Celcius
  • 25. What type of mathematical decay does the temperature of a cooling body follow?
A) Linear decay
B) Quadratic decay
C) Exponential decay
D) Logarithmic growth
  • 26. To solve for time t from the cooling formula, which mathematical operation must be used to eliminate the exponential base e?
A) Integration by parts
B) Square root
C) Natural Logarithm ln
D) Common Logarithm
  • 27. If the initial temperature of a rubber sample matches the room temperature T(o) - T(s), What is T(t) after 50 mins?
A) It drops to absolute zero
B) At zero degrees
C) It equals T(s)
D) It doubles its value
  • 28. For a given system,`if e-kt = 0.25 after 5 mins, what will be the value of e-kt after 10 minutes?
A) 0.050
B) 0.125
C) 0.500
D) 0.0625
  • 29. Given that T(t) = 30 + 70e-kt. What was the initial temperature (T (o)) of the substance at t=0?
A) 30 degree Celcius
B) 40 degree Celcius
C) 100 degree Celcius
D) 70 degree Celcius
  • 30. Using the same equation T(t) = 30 + 70 e-0.2t, what is the final temperature the substance reaches after a long time?
A) 0 degree Celcius
B) 100 degree Celcius
C) 70 degree Celcius
D) 30 degree Celcius
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