Physics Questions and Answers – Thermal Properties of Matter – Calorimetry

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This set of Physics Multiple Choice Questions & Answers (MCQs) focuses on “Thermal Properties of Matter – Calorimetry”.

1. A calorimeter does not involve measurement of temperature. True or False?
a) True
b) False
View Answer

Answer: b
Explanation: A calorimeter is a device which measures heat flow. It calculates the temperature change and along with the knowledge of mass and specific heat, it we can calculate heat by using the formula:
Q = msΔT.
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2. What is the principle of a calorimeter?
a) Heat loss rate is directly proportional to temperature difference
b) The total heat lost by a hot body is same as the total heat gained by the cold body provided no heat is lost to the surroundings
c) Heat loss depends only on specific heat of a substance
d) For the same amount of heat flow from a sample to two different samples of same mass is a product of their specific heats and temperature difference
View Answer

Answer: b
Explanation: The principle of calorimeter states that heat lost by a hot body is the same as the heat gained by the cold body provided no heat is lost to the surroundings. The other statement that states that the amount of heat flow from a sample to two different samples of same mass is a product of their specific heats and temperature difference is true but it is not the principle of calorimetry. It is just a mathematical way of understanding heat transfer.

3. What is the definition of 1 calorie?
a) It is the heat required to raise the temperature of 1g of water from 14.5°C to 15.5°C at 760mm Hg
b) It is the heat required to raise the temperature of 1g of any substance by 14.5°C to 15.5°C at 760mm Hg
c) It is the heat required to raise temperature of 1g of water by 1°C at 760mm Hg
d) It corresponds to the heat supplied at 760mm Hg for 1°C raise in temperature
View Answer

Answer: a
Explanation: A calorie is the unit of energy and is defined as the amount of heat required to raise the temperature of 1g of water from 14.5°C to 15.5°C at 760mm Hg.
Its value is 4.18 joules.
Note that the calories we read behind food packets correspond to the amount of heat that quantity of food provides to our body which it uses for functioning.
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4. The temperature of 50g of water has to be risen from 15°C to 85°C by using steam at 120°C. What is the amount of steam, in gms, required for the same? Sw= 1cal/ g°C, Lvap = 540cal/g, Ssteam =0.48cal/g°C.
a) 6.29g
b) 8.34g
c) 7.22g
d) 6.19g
View Answer

Answer: d
Explanation: Heat lost by steam = Heat gained by water. Let ‘m’ be the mass of steam required.
∴ mSsteam (120-100) + mLvap + mSw (100-85) = 50Sw (85-15) The only variable is ‘m’.
On calculating it from the above equation, we get:
m = 6.19g.

5. 30g water at 30°C is mixed with 1g of steam at 100℃. Calculate the final temperature of the mixture. Sw = 1cal/ g°C, Lvap= 540cal/g.
a) 49.67°C
b) 38.33°C
c) 53.67°C
d) 41.33°C
View Answer

Answer: a
Explanation: 30*1*(T-30) = 1*540 + 1*1*(100-T)

∴ 30T – 900 = 540 + 100 – T

⇒ 31T = 1540

⇒ T = 49.67°C
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6. What should be the temperature of 1g of ice, so that when it’s added to 9g of water, lowers the temperature of water by 10°C? Sw = 1cal/ g°C, Sice = 0.5cal/ g°C, Lfus= 80cal/g.
a) 20°C
b) -20°C
c) 0°C
d) 10°C
View Answer

Answer: b
Explanation: Let the initial temperature of ice be T°C.
Heat lost by ice will be the same as the heat lost by water.

∴ 1*0.5*(0-T) + 1*80 = 9*1*10
⇒ T = -20°C.

7. Same amount of heat is added to two substances of same mass having specific heat capacities s1 & s2 such that s1 > s2. Select the correct statement.
a) The change in temperature of 1 is greater than the change in temperature of 2
b) The change in temperature of 1 is less than the change in temperature of 2
c) Both will have the same change in temperature
d) Temperature changes can’t be related with specific heat capacities
View Answer

Answer: b
Explanation: Q = msΔT,
so for the same mass and heat supplied, the product of ‘s’ and ΔT must be the same.
Therefore more the value of ‘s’, less will be the value of ΔT.
Hence, ΔT will be lesser for 1 as ‘s’ is greater for 1 than 2.
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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter