NEET-XII-Physics
25: Calorimetry
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- Qstn #14A block of mass 100 g slides on a rough horizontal surface. If the speed of the block decreases from 10 m s-1 to 5 m s-1, find the thermal energy developed in the process.Ans : Given:
Mass of the block = 100 g = 0.1 kg
Initial speed of the block = 10 m/s
Final speed of the block = 5 m/s
Initial kinetic energy of the block = `` \frac{1}{2}\times 0.1\times {10}^{2}=5\,\mathrm{\,J\,}``
Final kinetic energy of the block = `` \frac{1}{2}\times 0.1\times {5}^{2}=1.25\,\mathrm{\,J\,}``
Change in kinetic energy of the block = 5 `` -`` 1.25 = 3.75 J
Thermal energy developed is equal to the change in kinetic energy of the block. Thus,
Thermal energy developed in the process = 3.75 J
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- Qstn #15The blocks of masses 10 kg and 20 kg moving at speeds of 10 m s-1 and 20 m s-1 respectively in opposite directions, approach each other and collide. If the collision is completely inelastic, find the thermal energy developed in the process.Ans : Given:
Mass of the first block, m1 = 10 kg
Mass of the second block, m2 = 20 kg
Initial velocity of the first block, u1 = 10 m/s
Initial velocity of the second block, u2 = 20 m/s
Let the velocity of the blocks after collision be v.
Applying conservation of momentum, we get
m2u2 - m1u1 = (m1 + m2)v
⇒ 20 × 20 - 10 × 10 = (10 + 20)v
⇒ 400 - 100 = 30 v
⇒ 300 = 30 v
⇒ v = 10 m/s
Initial kinetic energy is given by
`` {K}_{\,\mathrm{\,i\,}}=\frac{1}{2}{m}_{1}{u}_{1}^{2}+\frac{1}{2}{m}_{2}{u}_{2}^{2}``
`` {K}_{\,\mathrm{\,i\,}}=\frac{1}{2}\times 10\times {\left(10\right)}^{2}+\frac{1}{2}\times 20\times {\left(20\right)}^{2}``
`` {K}_{\,\mathrm{\,i\,}}=500+4000=4500``
Final kinetic energy is given by
`` {K}_{\,\mathrm{\,f\,}}\mathit{=}\frac{1}{2}\left({m}_{1}\mathit{+}{m}_{2}\right){v}^{2}``
`` {K}_{\,\mathrm{\,f\,}}=\frac{1}{2}\left(10+20\right){\left(10\right)}^{2}``
`` {K}_{\,\mathrm{\,f\,}}=\left(\frac{30}{2}\right)\times 100=1500``
∴ Total change in KE = 4500 J - 1500 J = 3000 J
Thermal energy developed in the process = 3000 J
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- Qstn #16A ball is dropped on a floor from a height of 2.0 m. After the collision it rises up to a height of 1.5 m. Assume that 40% of the mechanical energy lost goes as thermal energy into the ball. Calculate the rise in the temperature of the ball in the collision. Heat capacity of the ball is 800 J K-1.Ans : Height of the floor from which ball is dropped, h1 = 2.0 m
Height to which the ball rises after collision, h2 = 1.5 m
Let the mass of ball be m kg.
Let the speed of the ball when it falls from h1 and h2 be v1 and v2, respectively.
`` {v}_{1}=\sqrt{2g{h}_{1}}=\sqrt{2\times 10\times 2}=\sqrt{40}\,\mathrm{\,m\,}/\,\mathrm{\,s\,}``
`` {v}_{2}=\sqrt{2g{h}_{2}}=\sqrt{2\times 10\times 1.5}=\sqrt{30}\,\mathrm{\,m\,}/\,\mathrm{\,s\,}``
Change in kinetic energy is given by
`` ∆K=\frac{1}{2}\times \,\mathrm{\,m\,}\times 40-\left(\frac{1}{2}\,\mathrm{\,m\,}\right)\times 30=\left(\frac{10}{2}\right)\,\mathrm{\,m\,}``
`` \Rightarrow ∆K=5\,\mathrm{\,m\,}``
If the position of the ball is considered just before hitting the ground and after its first collision, then 40% of the change in its KE will give the change in thermal energy of the ball. At these positions, the PE of the ball is same. Thus,
Loss in PE = 0
The change in kinetic energy is utilised in increasing the temperature of the ball.
Let the change in temperature be ΔT. Then,
`` \left(\frac{40}{100}\right)\times ∆K=m\times 800\times ∆T``
`` \left(\frac{40}{100}\right)\times \frac{10}{2}m=m\times 800\times ∆T``
`` \Rightarrow ∆T=\frac{1}{400}=0.0025``
`` =2.5\times {10}^{-3}°\,\mathrm{\,C\,}``
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- Qstn #17A copper cube of mass 200 g slides down on a rough inclined plane of inclination 37° at a constant speed. Assume that any loss in mechanical energy goes into the copper block as thermal energy. Find the increase in the temperature of the block as it slides down through 60 cm. Specific heat capacity of copper = 420 J kg-1 K-1.Ans : Mass of copper cube, m = 200 g = 0.2 kg
Length through which the block has slided, l = 60 cm = 0.6 m
Since the block is moving with constant velocity, the net force on it is zero. Thus,
Force of friction, f = mg
Also, since the object is moving with a constant velocity, change in its K.E will be zero.As the object slides down, its PE decreases at the cost of increase in thermal energy of copper.
The loss in mechanical energy of the copper block = Work done by the frictional force on the copper block to a distanceof 60 cm
W = mg l sin θ
W = 0.2 × 10 × 0.6 sin 37°
W `` =1.2\times \left(\frac{3}{5}\right)=0.72``
Let the change in temperature of the block be ∆T.
Thermal energy gained by block = ms ∆T = 0.2 × 420×∆T = 84∆T
But 84∆T = 0.72
`` \Rightarrow ∆T=\frac{0.72}{84}=0.00857``
`` ∆T=0.0086=8.6\times {10}^{-3}°\,\mathrm{\,C\,}``
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- Qstn #18A metal block of density 600 kg m-3 and mass 1.2 kg is suspended through a spring of spring constant 200 N m-1. The spring-block system is dipped in water kept in a vessel. The water has a mass of 260 g and the bloc is at a height 40 cm above the bottom of the vessel. If the support of the spring is broken, what will be the rise in the temperature of the water. Specific heat capacity of the block is 250 J kg-3 K-1 and that of water is 4200 J kg-1 K-1. Heat capacities of the vessel and the spring are negligible.Ans : Given:
Density of metal block, d = 600 kg m-3
Mass of metal block, m = 1.2 kg
Spring constant of the spring, k = 200 N m-1
Volume of the block, V`` =\frac{1.2}{6000}=2\times {10}^{-4}{\,\mathrm{\,m\,}}^{3}``
When the mass is dipped in water, it experiences a buoyant force and in the spring there is potential energy stored in it.
If the net force on the block is zero before breaking of the support of the spring, then
kx + Vρg = mg
200x + (2 × 10-4)× (1000) × (10) = 12
`` \Rightarrow x=\frac{\left(12-2\right)}{200}``
`` \Rightarrow x=\frac{10}{200}=0.05\,\mathrm{\,m\,}``
The mechanical energy of the block is transferred to both block and water. Let the rise in temperature of the block and the water be ΔT.
Applying conservation of energy, we get
`` \frac{1}{2}k{x}^{2}+mgh-V\,\mathrm{\,\rho \,}gh={m}_{1}{s}_{1}∆T+{m}_{2}{s}_{2}∆T``
`` \Rightarrow \frac{1}{2}\times 200\times 0.0025+1.2\times 10\times \left(\frac{40}{100}\right)-2\times {10}^{-4}\times 1000\times 10\times \left(\frac{40}{100}\right)``
`` =\left(\frac{260}{1000}\right)\times 4200\times ∆T+1.2\times 250\times ∆T``
`` \Rightarrow 0.25+4.8-0.8=1092∆T+300∆T``
`` \Rightarrow 1392∆T=4.25``
`` \Rightarrow ∆T=\frac{4.25}{1392}=0.0030531``
`` ∆T=3\times {10}^{-3}°\,\mathrm{\,C\,}``