Tag: temperature and heat

Questions Related to temperature and heat

Two thermometers are constructed in the same way except that one has a spherical bulb and the other a cylindrical bulb; which one will respond quickly to temperature changes?

  1. Spherical bulb thermometer

  2. Cylindrical bulb thermometer

  3. both equally

  4. either


Correct Option: A

At the top of a mountain, a thermometer read ${7}^{0}C$ and barometer reads $70\, cm$ of Hg. At the bottom of the mountain, the barometer reads $76\,cm$ of Hg and thermometer reads ${27}^{0}C$. The density the density at the bottom.

  1. $0.99$

  2. $0.9$

  3. $0.89$

  4. $0.95$


Correct Option: A

A thermometer reads $0^{\circ}$ as $10^{\circ}$ and $100^{\circ}$ as $90^{\circ}C$. Then
Find the reading of the temperature if the temperature is $40^{\circ}C$?

  1. $42^{\circ}C$

  2. $40^{\circ}C$

  3. $35^{\circ}C$

  4. $45^{\circ}C$


Correct Option: A

If two temperatures differ by 25 degrees on Celsius scale, the difference of temperature on Fahrenheit scale is 

  1. $65^{\circ}$

  2. $45^{\circ}$

  3. $38^{\circ}$

  4. $25^{\circ}$


Correct Option: B
Explanation:

Each Celsius degree equals 5/9ths of a Fahrenheit degree. 

Each Fahrenheit degree equals 9/5ths of a Celsius degree. 

A 25 degree difference in Celsius would be 25 x 9/5 = 45 degrees Fahrenheit.

A Centigrade and Fahrenheit thermometers are dipped in boiling water. The water temperature is lowered unit the Fahrenheit thermometre registers a temperature of $140^o$. The fall of temperature as registered by the Centigrade thermometre is :

  1. $80^o$C

  2. $40^o$C

  3. $50^o$C

  4. $90^o$C


Correct Option: B

On a new scale of temperature (which is linear) and called the W scale, the freezing and boiling points of water are $39^o$W and $239^o$W respectively. What will be the temperature on the new scale, corresponding to a temperature of $39^o$C on the Celsius scale ?

  1. $78^o$W

  2. $117^o$W

  3. $200^o$W

  4. $139^o$W


Correct Option: B

A 1200 kg car is braked to a stop from 100 km/h. If 60% of this energy appears in the steel brake drums, whose total mass is 10 kg its temperature will rise by $\left( s _ { \text { steel } } = 450 \mathrm { J } / \mathrm { kg } \mathrm { K } \right)$

  1. $31.2 ^ { \circ } C$

  2. $61.7 ^ { \circ } \mathrm { C }$

  3. $92.1 ^ { \circ } C$

  4. $42.1 ^ { \circ } C$


Correct Option: B
Explanation:

$\large \begin{array}{l} According\, to\, question............. \ Here, \ \, \, \, \, \, m=1200kg,\, \, \, break\, drum\, (m)=10kg \ \, \, \, \, u=100km/h=27.77m/s \ \, \, \, \, { S _{ Steel } }=450J/KgK,\, \, \, \, \, V=0\, m/s \ so,\,  \ Energy\, released\, during\, breaking=\, \, change\, in\, kinetic\, energy \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\frac { 1 }{ 2 } m{ v^{ 2 } }-\frac { 1 }{ 2 } m{ u^{ 2 } } \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\frac { 1 }{ 2 } (1200)\, { (0)^{ 2 } }-\frac { 1 }{ 2 } (1200)\, { (27.77)^{ 2 } } \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =0-462937.0374\, J \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =462937.0374\, J \ Now, \ \, \, 60\quad perecent\quad of\, this\, energy\, appears=0.6\times 462937.0374\, J=277762.225 \ For\, change\, in\, temperature:\, \, \, \, ms\Delta \frac { 1 }{ 2 } t=277762.225 \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \Rightarrow \Delta t=\dfrac { { 277762.225 } }{ { ms } } =\dfrac { { 277762.225 } }{ { 10\times 450 } }  \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \therefore \, \, \, \, \Delta t\, ={ 61.7^{ 0 } }C \ so\, \, the\, correct\, option\, is\, B. \end{array}$

$10$grams of hydrogen gas and $40$ grams of helium gas are mixed in a closed vessel. Heat that has to be supplied to the gas raise its temperature by $50^oC$ is

  1. $2750$J

  2. $2750$cal

  3. $2000$cal

  4. $2000$J


Correct Option: A

To mark a temperature scale on a thermometer, standard temperatures knows as fixed points are needed.
Which of these is a fixed point on the Celsius scale?

  1. room temperature

  2. the temperature inside a freezer

  3. the temperature of pure melting ice

  4. the temperature of pure warm water


Correct Option: C
Explanation:

To define the Celsius temperature scale two "reliable" temperatures are used. These are, the melting point of pure ice, which gives us the lower fixed point of the scale and the boiling point of pure water (at atmospheric pressure) which gives us the upper fixed point.

A mercury thermometer, with a concave reflector behind the bulb, is placed in front of an electric fire. Which of the following combinations will cause the smallest reading on the thermometer?

  1. Black reflector, black bulb

  2. Black reflector, shiny bulb

  3. Shiny reflector, shiny bulb

  4. Remains same


Correct Option: B
Explanation:

A mercury thermometer, with a concave reflector behind the bulb, is placed in front of an electric fire.Black reflector, shiny bulb combinations will cause the smallest reading on the thermometer The black coating absorbs energy and warms up and warms the bulb by conduction.