Tag: measurement and effects of heat

Questions Related to measurement and effects of heat

A man would feel iron and wooden balls equally cold or hot at

  1. $98.6^oC$

  2. $98.6^oF$

  3. $198.6^oF$

  4. $198.6^oC$


Correct Option: B
Explanation:

A man would find them equally cold or hot only when the heat flowing in or out of them is equal.
But since there would be the same temperature difference between the object and the man.
The heat flow would only be the same, when there is no heat flow. i.e. the body's temperature should be equal to the body temperature of the man. Which is $98.6 ^{\circ} F$

How many calories of heat are required by gram of water at $99^oC$ to boil off:

  1. 530

  2. 640

  3. 540

  4. 500


Correct Option: C
Explanation:

Heat $= 4.2(1)(1)+ 1(2260) = 2264.2 J$
$1 cal = 4.2J$
Thus, Heat $= 539.1 cal \approx 540 cal$

At which temperature do the readings of the celcius and the Fahrenheit scales coincide ?

  1. $0$

  2. $100$

  3. $-40$

  4. $-80$

  5. None of the above


Correct Option: C
Explanation:

We know the relation between celsius and Fahrenheit is given by , 

$ ^0 C = \dfrac{9}{5}  \times ^0F + 32 $
For readings to coincide ,  $ ^0 C = ^0 F$
$ ^0 C = \dfrac{9}{5}  \times ^0C + 32 $
On Solving, $ ^0 C = -40 $


The amount of heat required to convert 1 g of ice (specific 0.5 cal  at $g^{-1o} C^{-1}$ ) at $-10^0 C$ to steam at $100 $ $^\circ C$ is ___________.

[ Given: Latent heat of ice is $80 Cal/ gm,$ Latent heat of steam is $540 Cal/gm $, Specific heat of water is $1 Cal/gm/C$ ]

  1. 725 cal

  2. 636 cal

  3. 716 cal

  4. None of these


Correct Option: A
Explanation:

Amount of heat required = Change the temp Of Ice from -10 C to 0 C + Heat required to melt the Ice + Heat required to increase the temperature of water from 0 to 100 C + Heat required to convert water  at 100 C to vapor at 100 C

$=1\times 0.5[0-(-10)]+1\times 80+1\times 1\times 100+1\times 540\ =5+80+100+540\ =725cal  $

A mass of stainless steel spoon is 0.04 kg and specific heat is $0.50 kJ/kg \times ^oC$. Then calculate the heat which is required to raise the temperature $20^oC$ to $50^oC$ of the spoon.

  1. 200 J

  2. 400 J

  3. 600 J

  4. 800 J

  5. 1,000 J


Correct Option: C
Explanation:

The required heat, $Q=ms\Delta T=0.04\times (0.50\times 10^3)\times(50-20)=600 J$

A body having $1680 J$ of energy is supplied to $100 g$ of water. If the entire amount of energy is converted into heat the rise in temperature of water (sp. heat of water = $4200 JKg^{ -1 }\ ^0C ^{ -1 } $)

  1. $0.4^{ \circ }{ C }$

  2. $40^{ \circ }{ C }$

  3. $4^{ \circ }{ C }$

  4. $44^{ \circ }{ C }$


Correct Option: C
Explanation:

If $\delta T$ is the rise in temperature, then the amount of heat supplied is $Q=mS\Delta T$  where $S=$ specific heat

Thus, $1680=(100/1000)(4200)\Delta T$ or $\Delta T=4^oC$