## Heat is supplied to a certain homogeneous sample of matter at a uniform rate. Its temperature is plotted …

Q: Heat is supplied to a certain homogeneous sample of matter at a uniform rate. Its temperature is plotted against time as shown in the figure. Which of the following conclusions can be drawn ?

(A) its specific heat capacity is greater in the solid state than in the liquid state.

(B) its specific heat capacity is greater in the liquid state than in the solid state.

(C) its latent heat of vaporization is greater than its latent heat of fusion.

(D) its latent heat of vaporization is smaller than its latent heat of fusion

Ans: (A) , (C)

Solution:

Slope of graph is greater in the solid state i.e.,
temperature is rising faster, hence lower heat
capacity. The transition from solid to liquid state takes lesser
time, hence latent heat is smaller

## A container with 1 kg of water in it is kept in sunlight , which causes the water to get warmer than the surroundings .

Q: A container with 1 kg of water in it is kept in sunlight , which causes the water to get warmer than the surroundings . The average energy per unit time per unit area received due to the sunlight is 700 W/m2 and it is absorbed by the water over an effective area of 0.05 m2 . Assuming that the heat loss from the water to the surroundings is governed by Newton’s Law of cooling , the difference (in °C ) in the temperature of water and the surroundings after a long time will be ……….. (Ignore effect of the container and take constant for Newton’s Law of cooling = 0.001 s-1 , Heat capacity of water = 4200 JKg-1 K-1 )

Ans: (8.33)

Solution: $\displaystyle \frac{dQ}{dt} = e \sigma A(T^4 – T_o^2)$

For small temperature change ,

$\displaystyle \frac{dQ}{dt} = e \sigma A T^3 \Delta T$ …(i)

$\displaystyle \frac{mC dT}{dt} = e \sigma A T^3 \Delta T$

$\displaystyle \frac{dT}{dt} = \frac{e \sigma A T^3}{m C} \Delta T$

$\displaystyle \frac{e \sigma A T^3}{m C}$ -> Constant for Newton’s Law of cooling

$\displaystyle \frac{e \sigma A T^3}{m C} = 0.001$

$\displaystyle e \sigma A T^3 = m c \times 0.001 = 1 \times 4200 \times 0.001$

$\displaystyle e \sigma A T^3 = 4.2$ …(ii)

$\displaystyle \frac{dQ}{dt} = 700 \times 0.05 = 35 W$ ….(iii)

Putting the value of Eqs (ii) & (iii) in Eq.(i)

35 = 4.2 ΔT

ΔT = 35/4.2 = 8.33

## The filament of a light bulb has surface area 64 mm^2 . The filament can be considered as a black body …..

Q: The filament of a light bulb has surface area 64 mm2 . The filament can be considered as a black body at temperature 2500 K emitting radiation like point source when viewed from far . At night the light bulb is observed from a distance of 100 m . Assume the pupil of the eyes of the observer to be circular with radius 3 mm . Then
(Stefan’s Boltzmann constant = 5.67 x 10-8 W m-2 K-4 , Wien’s displacement constant = 2.90 x 10-3 m-K , PlancK ‘ s constant = 6.63 x 10-34 J-s , Speed of light in vacuum = 3 x 108 m/s )

(a) Power radiated by the filament is in the range 642 W to 645 W

(b) Radiated power entering into one eye of the observer is in the range 3.15 x 10-8 W to 3.25 x 10-8 W

(c) The wavelength corresponding to the maximum intensity of light is 1160 nm

(d) Taking the average wavelength of emitted radiation to be 1740 nm , the total number of photons entering per second into one eye of the observer is in the range 2.75 x 1011 to 2.85 x 1011 .

Ans: (b , c , d)

Solution: Let A = surface area of filament , T = temperature of filament , d = distance of bulb from observer and Re = pupil of eye

A = 64 mm2 , T = 2500 K , d = 100 m , Re = 3 mm

(a) $P = \sigma A e T^4$

$P = 5.67 \times 10^{-8} \times 64 \times 10^{-6} \times 1 \times (2500)^4$ (Since e = 1 foe black body)

P = 141.7 W

(b) Power reaching to the eye

$= \frac{P}{4 \pi d^2} \times (\pi R_e^2 )$

$\frac{141.7}{4 \pi (100)^2}\times \pi \times (3 \times 10^{-3})^2$

= 3.189375 W

(c) $\lambda_m T = b$

$\lambda_m \times 2500 = 2.9 \times 10^{-3}$

$\lambda_m = 1.16 \times 10^{-6}$

= 1160 nm

(d) Power received by one eye of observer

$= \frac{h c}{\lambda} \times \frac{\Delta N}{\Delta t}$

Where , $\frac{\Delta N}{\Delta t}$ = no. of photons entering eye per second

$3.189375 \times 10^{-8} = \frac{6.63 \times 10^{-34}\times 3 \times 10^8}{1740 \times 10^{-9}} \times \frac{\Delta N}{\Delta t}$

$\frac{\Delta N}{\Delta t} = 2.79 \times 10^{11}$

## A bakelite beaker has volume capacity of 500 cc at 30 °C .When it is partially filled with volume Vm at 30°C of mercury ….

Q: A bakelite beaker has volume capacity of 500 cc at 30 °C .When it is partially filled with volume Vm at 30°C of mercury . , it is found that the unfilled volume of the beaker remains constant as the temperature is varied . If γbeaker = 6 × 10-6/°C and γmercury = 1.5 × 10-4/°C , where γ is the coefficient of volume expansion , then Vm in cc is close to ……

Click to See Solution :
Ans: (20)

Sol: Given V = 500 cc , T = 30 °C

γbeaker = 6 × 10-6/°C

and γmercury = 1.5 × 10-4/°C

V – Vm = V’ – Vm

Now , V’ = V(1 + γb ΔT)

and , Vm‘ = Vm(1 + γm ΔT)

V – Vm = V + V γb ΔT – Vm – Vm γm ΔT)

$\displaystyle V_m = V \frac{\gamma_b}{\gamma_m}$

$\displaystyle V_m = 500 \times \frac{6 \times 10^{-6}}{1.5 \times 10^{-4}}$

Vm = 20 cc

## A non-isotropic solid metal cube has coefficients of linear expansion as 5 × 10-5 °/C  along the x -axis ….

Q: A non-isotropic solid metal cube has coefficients of linear expansion as 5 × 10-5 °/C  along the x -axis and 5 × 10-6 °/C  along the y -axis and z -axis . If the coefficient of volume expansion of solid is C × 10-5 °/C then the value of C is ….

Click to See Solution :
Ans: (60)

Sol: αlx = 5 × 10-5 °/C

αly = αlz = 5 × 10-6 °/C

αv = αlx + αly + αlz

αv = αlx + 2 αly

αv = 5 × 10-5 + 2 × 5 × 10-6

= 6 × 10-5 °/C

= 60 × 10-6 °/C

C = 60

## A bullet of mass 5 g , travelling with a speed of 210 m/s , strikes a fixed wooden target …..

Q: A bullet of mass 5 g , travelling with a speed of 210 m/s , strikes a fixed wooden target . One half of its kinetic energy is converted into heat in the bullet while the other half is converted into heat in the wood . The rise of temperature of the bullet if the specific heat of its material is 0.030 cal/g°C . (1 cal = 4.2 × 107 ergs)

(a) 87.5 °C

(b) 83.3 °C

(c) 119.2 °C

(d) 38.4 °C

Click to See Solution :
Ans: (a)
Sol: Given m = 5 g , v = 210 m/s

As W = J Q

$\displaystyle \frac{1}{2}\frac{1}{2}m v^2 = J \times m \times s \times \Delta T$

$\displaystyle \Delta T = \frac{v^2}{4 J s }$

$\displaystyle \Delta T = \frac{(210)^2}{4 J s}$

= 87.5 °C

## A leakproof cylinder of length 1 m , made of a metal which has very low coefficient of expansion is floating vertically ….

Q: A leakproof cylinder of length 1 m , made of a metal which has very low coefficient of expansion is floating vertically in water at 0°C such that its height above the water surface is 20 cm . When the temperature of water is increased to 4°C , the height of the cylinder above the water surface becomes 21 cm . The density of water at T = 4°C , relative to density at T = 0°C is close to

(a) 1.03

(b) 1.01

(c) 1.26

(d) 1.04

Click to See Solution :
Ans: (b)

Sol: Let A be the area of cross section of cylinder

As height of cylinder above water surface = 20 cm

height of cylinder inside the water surface = 80 cm

When water at 0°C ,

m g = 80 × A × ρ0°C × g …(i)

When water at 4°C ,

m g = 79 × A × ρ4°C × g …(ii)

From (i) & (ii)

$\displaystyle \frac{\rho_{4^oC}}{\rho_{0^oC}} = \frac{80}{79} = 1.01$

## Heat is being supplied at a constant rate to a sphere of ice which is melting at the rate of 0.1 g/sec . It melts completely in 100 sec …

Q: Heat is being supplied at a constant rate to a sphere of ice which is melting at the rate of 0.1 g/sec . It melts completely in 100 sec . The rate of rise of temperature thereafter will be (Assume no loss of heat)

(a) 0.8 °C/sec

(b) 5.4 °C/sec

(c) 3.6 °C/sec

(d) None

Ans: (a)

Sol: m/t = 0.1

m = 0.1 t = 0.1 × 100 = 10 gm

Q = mL

$\displaystyle \frac{dQ}{dt} = L\frac{dm}{dt}$

$\displaystyle \frac{dQ}{dt} = 80 \times 0.1$

= 8 cal/sec

Again , $\displaystyle \frac{dQ}{dt} = m s \frac{d\theta}{dt}$

$\displaystyle \frac{d\theta}{dt} = \frac{\frac{dQ}{dt}}{m s}$

$\displaystyle \frac{d\theta}{dt} = \frac{8}{10 \times 1}$

= 0.8

## 10 gram of ice at 0 °C is kept in a calorimeter of water equivalent 10 gm . How much heat should be supplied ….

Q: 10 gram of ice at 0 °C is kept in a calorimeter of water equivalent 10 gm . How much heat should be supplied to the apparatus to evaporate the water thus formed ? (Neglect loss of heat)

(a) 6200 cal

(b) 7200 cal

(c) 13600 cal

(d) 8200 cal

Ans: (d)

Sol: Q = mcal scal (100 – 0) + mice[Lf + swater(100-0) + Lv]

= 10 × 1 × 100 + 10[80 + 1×100 + 540]

= 8200 cal

## A block of mass 2.5 kg is heated to temperature of 500 °C and placed on a large ice block . What is the maximum amount of ice …

Q: A block of mass 2.5 kg is heated to temperature of 500 °C and placed on a large ice block . What is the maximum amount of ice that can be melt (approx.) . Specific heat of the body = 0.1 cal/g°C

(a) 1 kg

(b) 1.5 kg

(c) 2 kg

(d) 2.5 kg

$\displaystyle m_{ice , melted} = \frac{2.5 \times 10^3 \times 0.1 \times 500}{80}$