Q:1. A 10 KW drilling machine is used to drill a bore in a small aluminium block of mass 8kg. How much is the rise in temperature of the block in 2.5 minutes assuming 50% of power is used up in heating the machine, specific heat of aluminium = 0.91 Jg-1C-1
Given, P = 10 KW = 104W, mass m = 8 kg = 8 × 103g,
t = 2.5 min = 2.5 × 60 = 150 s
Sp. heat, C = 0.91 Jg-1C-1
Total energy, P × t = 104 × 150
= 15 × 105J
Energy available, ∆Q =(1/2) ×15×105
= 7.5×105 (As 50% of energy is lost)
Again, Q = m c ∆T
∆T = ∆Q/mc
Q:2. A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500°C and then placed on a large ice block, what is the maximum amount of ice that can melt? Specific heat of copper is 0.397J/g-°C-1. Heat of fusion of water = 335 Jg-1
Given, Mass of copper block = m = 2.5 kg = 2500g;
Fall in temperature ∆T = 500 – 0 = 500°C
Specific heat of copper = C = 0.39 Jg-1°C-1;
Latent heat of fusion, L = 335 Jg-1
Let m’ = mass of ice melted
m’L = m c ∆T
m’ = mc∆T/L
Q:3. In an experiment on the specific hat of a metal, a 0.2 kg block of the metal at 150°C is dropped in copper calorimeter (or water equivalent 0.025 Kg) containing 150 cc of water at 27°C. The final temperature is 40°C. Calculate the specific heat of the metal. If heat losses to the surrounding are not negligible, is your answer greater or smaller than the actual value of specific heat of the metal ?
Given m = mass of metal = 0.2 kg = 200 g;
∆T = fall in temperature = 150 – 40 = 110°; c = specific heat of the metal
Heat lost by metal, ∆Q = mc∆T = 200 × c × 110 …(a)
Volume of water = 150 cc; Mass of water m= 150g;
Water equivalent of calorimeter, W = 0.025 kg = 25g
Rise in temperature of water and calorimeter,
∆T’ = 40 – 27 = 13°C
Heat gained by water and calorimeter
∆Q’ = (m’ + W)∆T’
= (150 + 25) × 13 = 175 × 13 …(b)
Since, ∆Q = ∆Q’
200 × c × 110 = 175 × 13 (from a and b)
c = 0.1 J/g/°C
Q:4.A gas heats water flowing at the rate of 3.0 liter/min from 27°C to 77°C. If the geyser operates on a gas burner, what is the rate of combustion of fuel, if its heat of combustion is 4 × 104 J/g ?
Sol: Given, volume of water heated = 3 liter/min,
Mass of water heated, m = 3000 g/min
Rise in temperature, ∆T = 77 – 27 = 50°C, Specific heat of water, c = 4.2 Jg-1°C-1
Amount of heat used, ∆Q = m c ∆T = 3000 × 4.2 × 50 = 63 × 104 J/min
Heat of combustion = 4 × 104 J/g ;
Rate of combustion of fuel =(63×104)/(4×10104 )
Q:5. What amount of heat must be supplied to 2 × 10-2 kg of nitrogen at room temperature to raise its temperature by 45°C at constant pressure? Given molecular mass of nitrogen is 28 and R = 8.3 J mol-1K-1.
Given, m = mass of gas = 2 × 10-2 kg = 20 g; Rise in temperature ∆T = 45°C
Molecular mass, M = 28
Heat required, ∆Q = n Cp ∆T
= (m/M)Cp ∆T
Q:6. Explain why
(a)Two bodies at different temperatures T1 and T2 , if brought in thermal contact to not necessarily settle to the mean temperature (T1+T2)/2 ?
(b)The coolant in a chemical or nuclear plant (i.e. the liquid used to prevent different parts of a plant from getting too hot) should have high specific heat. Comment.
(c)Air pressure in a car type increases during driving , why?
(d)The climate of a harbor town has more temperature (i.e. without extremes of heat and cold) than that of a town a desert at the same latitude, why?
(a)In thermal contact, heat from the body at higher temperature transfers to the body at lower temperature, till temperature of both becomes equal. Final temperature can be mean temperature. (T1+T2)/2 only when thermal capacities of two bodies are equal.
(b)Because heat absorbed ∝ specific heat of substance.
(c)Temperature of air inside the tyre increases due to motion during driving. Air pressure therefore, increase inside the tyre (by Charle’s law, P ∝ T)
(d)In a harbor town, the relative humidity is more than in a desert town. Hence, the climate of a harbor town is without extreme of hot and cold.