Download App

Thermal Properties of Matter — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsThermal Properties of Matter5 Marks
Question
A $10kW$ drilling machine is used to drill a bore in a small aluminium block of mass $8.0\ kg$. 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 itself or lost to the surroundings. Specific heat of aluminium $= 0.91J g^{–1} K^{–1}.$
✓

Answer

Power of the drilling machine, $P =10 kW=10 \times 10^3 W$
Mass of the aluminum block, $m =8.0 kg=8 \times 10^3 g$
Time for which the machine is used, $t =2.5 min=2.5 \times 60=150 s$
Specific heat of aluminium, $c =0.91 Jg ^{-1} K^{-1}$
Rise in the temperature of the block after drilling $=\delta T$
Total energy of the drilling machine $= Pt$
$= 10 \times 10^3\times 150$
$= 1.5 \times 10^6J$
It is given that only 50% of the power is useful. Useful energy, $\triangle\text{Q}=\Big(\frac{50}{100}\Big)\times1.5\times10^6=7.5\times10^5\text{J}$
But $\triangle\text{Q}=\text{mc}\triangle\text{T}$
$\therefore\triangle\text{T}=\frac{\triangle\text{Q}}{\text{mc}}$
$=\frac{(7.5\times10^5)}{(8\times10^3\times0.91)}$
$=103^\circ\text{C}$
Therefore, in 2.5 minutes of drilling, the rise in the temperature of the block is 103°C.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "5 Marks Questions" from Thermal Properties of Matter→Thermal Properties of Matter — all question types→Physics STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

Two masses $8kg$ and $12kg$ are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses and the tension in the string when the masses are released.
→
$0.32g$ of oxygen is kept in a rigid container and is heated. Find the amount of heat needed to raise the temperature from $25°C$ to $35°C$. The molar heat capacity of oxygen at constant volume is $20J\ mole^{-1}K^{-1}$.
→
In the previous problem, if $15.0cm$ of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms? (Specific gravity of mercury = $13.6$)
→
A ball is kept at a height h above the surface of a heavy transparent sphere made of a material of refractive index $\mu.$ The radius of the sphere is R. At t = 0, the ball is dropped to fall normally on the sphere. Find the speed of the image formed as a function of time for $\text{t}<\sqrt{\frac{2\text{h}}{\text{g}}}.$ Consider only the image by a single refraction.
→
A uniform square plate S(side c) and a uniform rectangular plate R(sides b, a) have identical areas and masses:
Show that:
  1. $\frac{\text{I}_\text{xR}}{\text{I}_\text{xS}}<1$
  2. $\frac{\text{I}_\text{ys}}{\text{I}_\text{ys}}>1$
  3. $\frac{\text{I}_{2\text{R}}}{\text{I}_{2\text{s}}}>1$
→
The half-life of $^{199}Au$ is 2.7 days.
  1. Find the activity of a sample containing $1.00\mu\text{g}$ of $^{198}Au$.
  2. What will be the activity after $7$ days? Take the atomic weight of $^{198}Au$ to be $198g/mol$.
→
A spring compressed by 0.1m develops a restoring force 10N. A body of mass 4kg placed on it. Deduce
  1. The force constant of the spring.
  2. The depression of the spring under the weight of the body(take g = 10 N/ kg).
  3. The period of oscillation, the body is distributed and.
  4. Frequency of oscillation.
→
Two large metal sheets carry surface currents as shown in figure. The current through a strip of width dl is Kdl where K is a constant. Find the magnetic field at the points P, Q and R.
→
From a certain apparatus, the diffusion rate of hydrogen has an average value of $28.7 \mathrm{~cm}^3 \mathrm{~s}^{-1}$. The diffusion of another gas under the same conditions is measured to have an average rate of $7.2 \mathrm{~cm}^3 \mathrm{~s}^{-1}$. Identify the gas. [Hint: Use Graham's law of diffusion $\frac{\mathrm{R}_1}{\mathrm{R}_2}=\left(\frac{\mathrm{M}_2}{\mathrm{M}_1}\right)^{\frac{1}{2}}$, where $\mathrm{R}_1, \mathrm{R}_2$ are diffusion rates of gases 1 and 2 , and $\mathrm{M}_1$ and $\mathrm{M}_2$ their respective molecular masses. The law is a simple consequence of kinetic theory]
→
Find the equivalent resistance of the circuits shown in the figure. between the points a and b. Each resistor has resistance r.

→