ThatQuiz Test Library Take this test now
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 mass
C) its specific heat capacity
D) The temperature difference between the body and the surroundings
  • 2. In Newton’s law of cooling, T s represents:
A) surrounding temperature
B) standard temperature
C) surface area
D) steam temperature
  • 3. Newton’s law of cooling is most accurate for
A) Melting processes
B) small temperature differences
C) large 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) vacuum surroundings
B) lower emissivity
C) smaller surface area
D) Larger temperature difference
  • 6. A correction in calorimetry is needed because
A) Thermometers are inaccurate
B) Water changes mass
C) Heat is lost to the surrounding
D) Heat is gained from the surroundings
  • 7. A liquid cools faster when
A) Mass increases
B) Air movement 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) 150 J/s
B) 2.5 J/s
C) 1.5 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) 1°C/s
B) 2°C/s
C) 4°C/s
D) 3°C/s
  • 11. The purpose of cooling correction is to obtain the:
A) Lowest temperature
B) Mass of the calorimeter
C) Room temperature
D) Actual quantity of heat exchanged
  • 12. Cooling correction is applied in calorimetry because
A) The calorimeter changes colour
B) Water evaporates completely
C) Heat is lost to the surroundings during the experiment
D) The thermometer expands
  • 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) Radiation and convection
B) Magnetism
C) Reflection
D) Compression
  • 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) 5%
D) 15%
  • 16. A temperature rise recorded is 15°C. If the cooling correction is +1°C, calculate the percentage correction.
A) Quantity of Heat accurately
B) Density
C) Volume
D) Pressure
  • 17. During a calorimetry experiment, the observed temperature rise is smaller than the true value because:
A) Temperature becomes constant immediately
B) Heat escapes to the surrounding
C) Water gains mass
D) The calorimeter melts
  • 18. The unit of quantity of heat is
A) Watt
B) Joule
C) Pascal
D) Newton
  • 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) 48.0°C
C) 47.7°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) 25°C
B) 30°C
C) 45°C
D) 35°C
  • 21. In the formula, what does the variable T(s) represent?
A) The time taken to reach thermal equilibrium
B) The constant temperature of the surrounding environment (surroundings)
C) The specific heat capacity of the material
D) The starting temperature of the object
  • 22. What does the variable T(t) represent in the equation
A) The total heat energy lost by the system
B) The temperature of the object at any specific time t
C) The time it takes for the temperature to drop to 0 degrees Celsius
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) 25 degree Celcius
C) 205 degree Celcius
D) 180 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) Quadratic decay
B) Linear 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) Common Logarithm
D) Natural Logarithm ln
  • 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 doubles its value
B) At zero degrees
C) It drops to absolute zero
D) It equals T(s)
  • 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) 0.0625
B) 0.500
C) 0.050
D) 0.125
  • 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) 100 degree Celcius
C) 40 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) 70 degree Celcius
C) 30 degree Celcius
D) 100 degree Celcius
Created with That Quiz — the math test generation site with resources for other subject areas.