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