Thermal Properties of Matter

5 NEET-UG practice questions on Thermal Properties of Matter, part of Physics. The first 5 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

1. Question 1 · Hard

   In a room at 20 °C, a body cools from 70 °C to 50 °C in 5 minutes. Assuming Newton's law of cooling, what is its temperature after a further 5 minutes?

   1. A40 °C
   2. B35 °C
   3. C38 °C
   4. D30 °C

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   Answer: C · 38 °C

   Using average temperatures: 20/5 = k(60 − 20), so k = 0.1 per minute. Then (50 − T)/5 = 0.1[(50 + T)/2 − 20], which gives T = 38 °C. Check: the excess over the room falls by the same ratio in equal times, 50 → 30 → 18, and 20 + 18 = 38 °C.

   • A: This assumes the drop simply halves in the second interval.
   • B: This uses the rate at 50 °C throughout, not the mean temperature.
   • D: This assumes the same 20 °C drop in each interval.

   Remember this

   Under Newton's law, the excess temperature falls by equal ratios, not equal amounts, in equal times.

2. Question 2 · Easy

   An isotropic solid has a coefficient of linear expansion α. What are its coefficients of area expansion and volume expansion, in that order?

   1. A2α and 3α
   2. B3α and 2α
   3. Cα/2 and α/3
   4. Dα² and α³

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   Answer: A · 2α and 3α

   For small temperature changes, an area grows through two lengths and a volume through three, so β = 2α and γ = 3α.

   • B: This interchanges the area and volume coefficients.
   • C: This divides by the dimension number instead of multiplying.
   • D: This raises α to a power; small changes add instead.

   Remember this

   α : β : γ = 1 : 2 : 3 for an isotropic solid.

3. Question 3 · Medium

   In a calorimetry experiment, 10 g of ice at 0 °C is added to 100 g of water at 30 °C in an insulated container of negligible heat capacity. Taking the latent heat of fusion of ice as 80 cal/g and the specific heat of water as 1 cal/(g °C), what is the final temperature?

   1. A0 °C
   2. B27.3 °C
   3. C22 °C
   4. D20 °C

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   Answer: D · 20 °C

   Heat lost by the water = heat to melt the ice + heat to warm the melted ice: 100(30 − T) = 10 × 80 + 10T. So 3000 − 800 = 110T and T = 20 °C. Check: the water cools by 10 °C and gives up 1000 cal, of which 800 cal melts the ice and 200 cal warms the 10 g by 20 °C.

   • A: This assumes not all the ice melts, although enough heat is available.
   • B: This ignores the latent heat of melting.
   • C: This forgets to warm the melted ice to the final temperature.

   Remember this

   Melted ice is water at 0 °C, which must still warm up to the final temperature.

4. Question 4 · Medium

   A 2 kg solid sample is heated steadily, and its temperature is plotted against heat supplied. What is the latent heat of fusion of the material?

   

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   1. A20 kJ/kg
   2. B120 kJ/kg
   3. C200 kJ/kg
   4. D100 kJ/kg

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   Answer: D · 100 kJ/kg

   Melting is the flat part, from 40 kJ to 240 kJ, so 200 kJ melts 2 kg. Therefore L = 200/2 = 100 kJ/kg.

   • A: Uses the heat of the sloping solid part, 40 kJ.
   • B: Reads the end of the plateau, 240 kJ, instead of its length.
   • C: Leaves out division by the mass.

   Remember this

   Latent heat is the length of the plateau on the heat axis divided by the mass.

5. Question 5 · Medium

   Two slabs of equal thickness L and equal area are joined face to face: conductivity k from x = 0 to L and 2k from x = L to 2L. In steady state the outer faces stay at 100 °C (x = 0) and 40 °C (x = 2L). Which graph shows temperature against x?

   

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   1. AGraph A
   2. BGraph D
   3. CGraph C
   4. DGraph B

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   Answer: C · Graph C

   Equal heat current gives k(100 − T)/L = 2k(T − 40)/L, so T = 60 °C at the joint. Temperature is linear in each slab, with a 40 °C drop in the first and a 20 °C drop in the second, as in Graph C.

   • A: Graph A assumes equal conductivities, with 70 °C at the joint.
   • B: Graph D is curved, but each uniform slab has a straight profile.
   • D: Graph B gives the larger drop to the better conductor.

   Remember this

   In series, heat current is common, so the poorer conductor takes the larger temperature drop.