Thermal Properties of Matter - Online Test

# Thermal Properties of Matter # CBSE 11TH PHYSICS Prepare / Learn

Q1.

A 10 kW drilling machine is used to drill a bore in a small aluminum block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50 percent of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminum = 0.91 J $g^{- 1} K^{- 1}$ .

A. 93

^{\circ}

B. 123

^{\circ}

C. 113

^{\circ}

D. 103

^{\circ}

Answer : Option D
Explaination / Solution:

$W = P t = 10 \times 10^{4} \times 2.5 \times 60 = 150 \times 10^{5} J$

Energy used in heating = 50% =75 x 105J

$\begin{aligned} Q = m s Δ T \\ Δ T = \frac{Q}{m s} \end{aligned}$

$Δ T = \frac{75 \times 10^{5}}{8000 \times 0.91}$

$Δ T = 103^{\circ} C$

# Thermal Properties of Matter # CBSE 11TH PHYSICS Prepare / Learn

Q2.

A child running a temperature of 101 $^{\circ}$ F is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98 $^{\circ}$ F in 20 min, what is the average rate of extra evaporation caused, by the drug? Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal $g^{- 1}$

g^{- 1}

A. 5.2 g/min

B. 4.6 g/min

C. 4.3 g/min

D. 4.9 g/min

Answer : Option C
Explaination / Solution:

$Δ T = 101 - 98 = 3^{\circ} F = {(3 \times \frac{5}{9})}^{\circ} C$

Specific heat of the human body = Specific heat of water = 1000 cal/kg/ °C

The heat lost by the child is $Q = m c Δ T = 30 \times 1000 \times (3 \times \frac{5}{9}) = 50000 c a l$

Let m1 be the mass of the water evaporated from the child’s body in 20 min.
Loss of heat through water is

$\begin{aligned} Q = m_{1} L \\ m_{1} = \frac{Q}{L} = \frac{50000}{580} = 86.2 g m \end{aligned}$

Average rate of extra evaporation caused by the drug = m1 / t
= 86.2 / 200
= 4.3 g/min.

# Thermal Properties of Matter # CBSE 11TH PHYSICS Prepare / Learn

Q3. A ‘thermacole’ icebox is a cheap and efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45

^{\circ}

C, and co-efficient of thermal conductivity of thermacole is 0.01 J

s^{- 1} m^{- 1} K^{- 1}

. [Heat of fusion of water = 335

\times

1 0^{3} J k g^{- 1}

]

A. 3.5 kg

B. 3.4 kg

C. 3.6 kg

D. 3.7 kg

Answer : Option D
Explaination / Solution:

$Q = \frac{K A Δ T}{l} t$

A = Surface area of the box = 6s2 = 6 × (0.3)2 = 0.54 m3

Time gap, t = 6 h = 6 × 60 × 60 s

$Q = \frac{0.01 \times 0.54 \times 45 \times 6 \times 60 \times 60}{0.05} = 104976 J$

But Q = m'L
∴ m' = Q/L
= 104976/(335 × 103) = 0.313 kg
Mass of ice left = 4 – 0.3 = 3.7 kg
Hence, the amount of ice remaining after 6 h is 3.7 kg.

# Thermal Properties of Matter # CBSE 11TH PHYSICS Prepare / Learn

Q4.

A brass boiler has a base area of 0.15 m2 and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 J $s^{- 1} m^{- 1} K^{- 1}$ ; Heat of vaporization of water = 2256 $\times$ $1 0^{3} J k g^{- 1}$

A. 237

^{\circ}

B. 239

^{\circ}

C. 240

^{\circ}

D. 238

^{\circ}

Answer : Option D
Explaination / Solution:

Base area of the boiler, A = 0.15 m2
Thickness of the boiler, l = 1.0 cm = 0.01 m
Boiling rate of water, R = 6.0 kg/min
Mass, m = 6 kg
Time, t = 1 min = 60 s
Thermal conductivity of brass, K = 109 J s –1 m–1 K–1
Heat of vaporisation, L = 2256 × 103 J kg–1
The amount of heat flowing into water through the brass base of the boiler is

$Q = \frac{K A (T_{1} - T_{2})}{l} t$ ..........(1)

where,
T1 = Temperature of the flame in contact with the boiler
T2 = Boiling point of water = 100°C
Heat required for boiling the water

Q = mL ....… (ii)
Equating equations (i) and (ii), we get:
∴ mL = KA(T1 - T2) t / l
T1 - T2 = mLl / KAt
= 6 × 2256 × 103 × 0.01 / (109 × 0.15 × 60)
= 137.98 o C
Therefore, the temperature of the part of the flame in contact with the boiler is 237.98°C.

# Thermal Properties of Matter # CBSE 11TH PHYSICS Prepare / Learn

Q5. A steel railroad track has a length of 30.000 m when the temperature is 0.0

^{\circ}

C. What is its length when the temperature is 40.0

^{\circ}

α = 11 \times 10^{- 6}

(

^{\circ}

^{- 1}

A. 30.016 m

B. 30.013 m

C. 30.010 m

D. 30.008 m

Answer : Option B
Explaination / Solution:

$Δ l = α l Δ θ = 11 \times 10^{- 6} \times 30 \times 40 = 0.013 m$

Total length after expansion = 30 + 0.013 = 30.013m

# Thermal Properties of Matter # CBSE 11TH PHYSICS Prepare / Learn

Q6. A steel railroad track has a length of 30.000 m when the temperature is 0.0

^{\circ}

C.Suppose that the ends of the rail are rigidly clamped at 0.0

^{\circ}

C so that expansion is prevented. What is the thermal stress set up in the rail if its temperature is raised to 40.0

^{\circ}

α - s t e e l = 11 \times 10^{- 6}

A. 8.1 × 10⁷N/m²

B. 8.4 × 106 N/m2

C. 8.8 × 106 N/m2

D. 9.1 × 107 N/m2

Answer : Option C
Explaination / Solution:

Thermal Stress

= Y α Δ θ = 20 \times 10^{9} \times 11 \times 10^{- 6} \times 40 = 8.8 \times 10^{6}

# Thermal Properties of Matter # CBSE 11TH PHYSICS Prepare / Learn

Q7. An ideal gas occupies a volume of 100

c m^{3}

2 0^{\circ}

C and 100 Pa. Find the number of moles of gas in the container.

A. 4.30

\times

1 0^{- 6}

B. 4.80

\times

1 0^{- 6}

C. 4.10×10−6

D. 4.60

\times

1 0^{- 6}

Answer : Option C
Explaination / Solution:

number of moles according to gas equation

$n = \frac{P V}{R T} = \frac{100 \times 100 \times 10^{- 6}}{8.31 \times 293} = 4.10 \times 10^{- 6}$

# Thermal Properties of Matter # CBSE 11TH PHYSICS Prepare / Learn

Q8.

A spray can containing a propellant gas at twice atmospheric pressure (202 kPa) and having a volume of 125 $c m^{3}$ is at 22 $^{\circ}$ C. It is then tossed into an open fire. When the temperature of the gas in the can reaches 195 $^{\circ}$ C, what is the pressure inside the can? Assume any change in the volume of the can is negligible.

290 kPa

300 kPa

C. 320 kPa

D. 260 kPa

Answer : Option C
Explaination / Solution:

P∝TP1P2=T1T2P2=468×202295=320KPa

# Thermal Properties of Matter # CBSE 11TH PHYSICS Prepare / Learn

Q9. A student eats a dinner rated at 2000 Calories. He wishes to do an equivalent amount of work in the gymnasium by lifting a 50.0-kg barbell. How many times must he raise the barbell to expend this much energy? Assume that he raises the barbell 2.00 m each time he lifts it and that he regains no energy when he drops the barbell to the floor.

A. 8.74

\times

1 0^{3}

B. 8.34

\times

1 0^{3}

C. 8.14

\times

1 0^{3}

D. 8.54

\times

1 0^{3}

Answer : Option D
Explaination / Solution:

Total no.of times he lift barbell = Total energy / energy in lifting barbell once

$\begin{aligned} = \frac{Q}{m g h} \\ = \frac{2000 \times 4.2 \times 10^{3}}{50 \times 10 \times 2} = 4.2 \times 10^{3} t i m e s \end{aligned}$

# Thermal Properties of Matter # CBSE 11TH PHYSICS Prepare / Learn

Q10. In an insulated vessel, 250 g of ice at 0

^{\circ}

C is added to 600 g of water at 18.0

^{\circ}

C. How much ice remains when the system reaches equilibrium?

A. 87 g

B. 134 g

C. 114 g

D. 94 g

Answer : Option C
Explaination / Solution:
No Explaination.

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