Thermal Properties of Matter Test

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Thermal Properties of Matter Test - 47

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Question 1
1 / -0

Water of volume $$2L$$ in a closed container is heated with a coil of $$1kW$$. While water is heated, the container loses energy at a rate of $$60J/s$$. In how much time will the temperature of water rise from $${27}^{o}C$$ to $${77}^{o}C$$ (Specific heat of water is $$4.2kJ/kg$$ and that of the container is negligible)

A
$$8min$$ $$20s$$

B
$$6min$$ $$2s$$

C
$$7$$ min

D
$$14$$ min

Solution

$$\textbf{Given}:$$ Volume of water = 2L

$$\textbf{Solution}:$$
From question heat gained by water in 1 second = 1 KW - 160 J/s = 1000 J/s - 160 J/s = 840 J/s
Total heat required to raise the temperature of water(volume 2L) from $$27^o C$$ to $$77^o C$$
= $$m_{water} \times \text{specific heat} \times \Delta \theta$$
$$= 2 \times 10^3 \times 4.2 \times 50 [ Q_{mass} = density \times volume]$$
Thus
$$840 \times t = 2 \times 10^3 \times 4.2 \times 50$$
OR
$$t = 2 \times 10^3 \times 4.2 \times \dfrac{50}{840}$$
t = 500 seconds
t = 8 min 20 sec

$$\textbf{Hence A is the correct option}$$
Question 2
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Two monatomic ideal gases $$1$$ and $$2$$ of molecular masses $$m_1$$ and $$m_2$$ respectively are enclosed in separate containers kept at the same temperature. The ratio of the speed of sound in gas $$1$$ to that in gas $$2$$ is given by.

A
$$\sqrt{\dfrac{m_1}{m_2}}$$

B
$$\sqrt{\dfrac{m_2}{m_1}}$$

C
$$\dfrac{m_1}{m_2}$$

D
$$\dfrac{m_2}{m_1}$$

Solution

$$V_{rm}=\sqrt {\dfrac {rRT}{m}}$$
i.e $$vr\dfrac {1}{\sqrt m}$$
which given $$v-1\times \dfrac {1}{\sqrt {m_1}}$$ and $$v_2\times \dfrac {1}{\sqrt {m_1}}$$
$$\dfrac {v_1}{v_2}=\sqrt {\dfrac {m_2}{m_1}}$$
Question 3
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An ideal gas is initially at $$P_1$$, $$V_1$$ is expanded to $$P_2, V_2$$ and then compressed adiabatically to the same volume $$V_1$$ and pressure $$P_3$$. If W is the net work done by the gas in the complete process which of the following is true.

A
$$W > 0; P_3 > P_1$$

B
$$W < 0; P_3 > P_1$$

C
$$W > 0; P_3 < P_1$$

D
$$W < 0; P_3 < P_1$$

Solution

Since work is done by the gas as area under the graph is $$-ve$$ ( anticlockwise) ie energy is lost by the gas and $$w<0$$ when $$P_3>P_1$$.
Question 4
1 / -0

$$300$$ grams of water at $$25^o$$C is added to $$100$$ grams of ice at $$0^o$$C. The final temperature of the mixture is ________ $$^oC$$.

A
2

B
0

C
3

D
4

Solution

$$\textbf{Given}:$$ m = 300g at $$25^{o}C$$ for water,$$m = 100g$$ at $$0^{o}C$$ for ice

$$\textbf{Solution}:$$
The heat required for 100g of ice at $$0^{o}C$$ to change into Water at $$0^{o}C=mL=100\times 80\times 4.2=33,600J$$…(i)
The heat released by 300g of water at $$25^{o}C$$ to change its temperature to $$0^{0}C=mc \Delta T=300\times 4.2\times 25=31,500J$$….(ii)
Since the energy in eq.(ii) is less than of eq(i) therefore the final temperature will be $$0^{o}C $$

$$\textbf{Hence B is the correct option}$$
Question 5
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A double pane window used for insulating a room thermally from outside, consists of two glass sheets each of area $$1m^2$$ and thickness $$0.01$$m separated by a $$0.05$$m thick segment air space. In the steady state the room glass interface and the glass outdoor interface are at constant temperature of $$27^o$$C and $$0^o$$C respectively. Calculate the rate of heat flow through the window pane. Also find the temperature of other interfaces. Given thermal conductivities of glass and air are as $$0.8$$ and $$0.08$$ $$Wm^{-1}K^{-1}$$ respectively.

A
$$41.54$$W, $$46.48^oC, 0.52^oC$$.

B
$$41.54$$W, $$26.48^oC, 0.52^oC$$.

C
$$41.54$$W, $$36.48^oC, 0.52^oC$$.

D
$$41.54$$W, $$66.48^oC, 0.52^oC$$.

Solution

As $$\dfrac {dQ}{dt}=K\dfrac {A\Delta Q}{L}=\dfrac {\Delta Q}{R}$$
Required $$=\displaystyle \sum \dfrac {L}{KA}=\dfrac {1}{A} \left [\dfrac {0.01}{0.8}\times 2+\dfrac {0.05}{0.08}\right]$$
$$A=1\ m^2$$, Required $$=\dfrac {1}{40}+\dfrac {5}{8}=\dfrac {26}{40}$$
$$\dfrac {dQ}{dt}=\dfrac {\Delta Q}{R}= \dfrac {(27-0\times 40)}{26}=41.5\ W$$
Now $$41.5=0.8\times 1^3 \dfrac {27-Q_2}{0.01}$$ or $$Q_2=26.48^oC$$
$$41.5=\dfrac {0.8\times 1^3 (Q_2 -0)}{0.01}, Q_1=0.52^oC$$
$$\dfrac {dQ}{dt}=41.5\ W$$
$$Q_2=26.48^oC$$
$$Q_1=0.52^oC$$
$$(B)$$
Question 6
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A cylinder of mass $$1$$kg is given heat of $$20000$$J at atmospheric pressure. If initially temperature of cylinder is $$20^o$$C, find final temperature of the cylinder.

A
$$T_{final}=90^oC$$.

B
$$T_{final}=70^oC$$.

C
$$T_{final}=80^oC$$.

D
$$T_{final}=60^oC$$.
Question 7
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We have half a bucket (6l) of water at 20C. If we want water at 40C, how much steam at 100C should be added to it?

A
200 g

B
2000/9 g

C
2 kg

D
200/3 g

Solution

6000 x 1 x (40 - 20) = m x 540 + m x 1 x (100 - 40)
12 x $$10^4$$ = m x 600
m = 2 x $$10^2$$ = 200 gm
Question 8
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Out of three thermometers one is kept in contact with the mans skin, second in contact with cotton vest and polyester shift and third in between shirt and woolen coat. The reading of the thermometers are $$35^{o}C$$, $$32^{o}C$$, $$25^{o}C$$. The ratio of $$K$$ for cotton and polyester will be

A
$$7\ :\ 3$$

B
$$3\ :\ 7$$

C
$$10\ :\ 3$$

D
$$3\ :\ 10$$

Solution

Based on heat balance, the distribution of temperature is described by a second order differential equation.

$$\dfrac {d^2T}{dx^2} = -h' (T_o-T)$$

The negative sign indicates a temperature decrease with time.

$$T_o$$ is surrounding temperature.

Temperature decreases towards right so, the ratio of the temperature $$k$$ for cotton and polyester is

$$k=\dfrac {h'(35-32)}{h'(32-25)}=\dfrac {3}{7}$$

So, $$k=\dfrac {3}{7}$$.

Hence, Option $$B$$ is the correct answer.
Question 9
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Temperature of hot end and cold end of a rod, which is in steady state, are $$100^{\circ}C$$ and $$40^{\circ}C$$ respectively. The area of cross section of rod and its thermal conductivity are uniform. The temperature of rod at its mid point is :
(Assume no heat loss through lateral surface).

A
$$70^{\circ}C$$

B
$$60^{\circ}C$$

C
$$80^{\circ}C$$

D
None of these

Solution

We know,

$$\dfrac{dQ}{dt}=\dfrac{k\times A\times (T_2-T_1)}{l}$$

Suppose the temperature at midpoint is $$T^0 C$$,

Now,

$$\dfrac{dQ}{dt}=\dfrac{kA(T-40)}{l/2}=\dfrac{kA(100-T)}{l/2}$$

$$100-T=T-40\\ 2T=140 \\ T=70^0 C$$

Hence option $$\textbf A $$ is correct answer.
Question 10
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Rate of heat flow through cylindrical rod is $${Q}_{1}$$. Temperatures of ends of rod are $${T}_{1}$$ and $${T}_{2}$$. If all the linear dimensions of the rod become double and temperature difference remains same, its rate of heat flow is $${Q}_{2}$$, then

A
$${Q}_{1}=2{Q}_{2}$$

B
$${Q}_{2}=2{Q}_{1}$$

C
$${Q}_{2}=4{Q}_{1}$$

D
$${Q}_{1}=4{Q}_{2}$$

Solution

$$H = kA \dfrac {(Q_{1} - Q_{2})}{L}T$$
Rate of neat flow
$$Q = \dfrac {H}{T} = \dfrac {KA(Q_{1} - Q_{2})}{t}$$
$$Q_{1} \times \dfrac {A}{L}$$ Dimension of $$Area = A = (L^{2})$$ Dimension of distance $$= (L)$$
$$QQL$$
$$\Rightarrow \dfrac {Q_{1}}{Q_{2}} = \dfrac {L}{L_{2}} = 2$$
$$\Rightarrow Q = 2Q_{2}$$.

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Thermal Properties of Matter Test - 47

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Thermal Properties of Matter Test - 47

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SHARING IS CARING

Question 1

$$8min$$ $$20s$$

$$6min$$ $$2s$$

$$7$$ min

$$14$$ min

Solution

Question 2

$$\sqrt{\dfrac{m_1}{m_2}}$$

$$\sqrt{\dfrac{m_2}{m_1}}$$

$$\dfrac{m_1}{m_2}$$

$$\dfrac{m_2}{m_1}$$

Solution

Question 3

$$W > 0; P_3 > P_1$$

$$W < 0; P_3 > P_1$$

$$W > 0; P_3 < P_1$$

$$W < 0; P_3 < P_1$$

Solution

Question 4

2

0

3

4

Solution

Question 5

$$41.54$$W, $$46.48^oC, 0.52^oC$$.

$$41.54$$W, $$26.48^oC, 0.52^oC$$.

$$41.54$$W, $$36.48^oC, 0.52^oC$$.

$$41.54$$W, $$66.48^oC, 0.52^oC$$.

Solution

Question 6

$$T_{final}=90^oC$$.

$$T_{final}=70^oC$$.

$$T_{final}=80^oC$$.

$$T_{final}=60^oC$$.

Question 7

200 g

2000/9 g

2 kg

200/3 g

Solution

Question 8

$$7\ :\ 3$$

$$3\ :\ 7$$

$$10\ :\ 3$$

$$3\ :\ 10$$

Solution

Question 9

$$70^{\circ}C$$

$$60^{\circ}C$$

$$80^{\circ}C$$

None of these

Solution

Question 10

$${Q}_{1}=2{Q}_{2}$$

$${Q}_{2}=2{Q}_{1}$$

$${Q}_{2}=4{Q}_{1}$$

$${Q}_{1}=4{Q}_{2}$$

Solution

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