Tag: heat exchange

Questions Related to heat exchange

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

physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

  1. 530

  2. 640

  3. 540

  4. 500

Show answer
Correct Option: C
Explanation:

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

$50  g$ of ice at 0 C is mixed with $50  g$ of water at 20 C.The resultant temperature of the mixture would be

physics calorimetry heat exchange calorimeter measuring thermal quantities by the method of mixtures

  1. 10 C

  2. 0 C

  3. -10 C

  4. -35 C

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Correct Option: B
Explanation:

$50 g $ of Ice at $0^0C$ has a latent heat of

$ Q = m \times L = 50 \times 80 $ 

                    $ = 4000 cal $

Now for water to reach $0^{0}C$ without changing its state 

Heat released by water = $ mC _p  \Delta T$

                                      = $ 50 \times 1 \times 20  $

                                      = $1000 cal $

As the heat to be removed from water is less than the latent heat of $50g $ of ice, the resultant mixture stays at $0^0C $ temperature.

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

physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

  1. $0$

  2. $100$

  3. $-40$

  4. $-80$

  5. None of the above

Show answer
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 $1 Cal/gm/C$ ]

physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

  1. 725 cal

  2. 636 cal

  3. 716 cal

  4. None of these

Show answer
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  $

The specific heat for substance $A$ is twice the specific heat of substance $B$. The same mass of each substance is allowed to gain $50$ Joules of heat energy. As a result of the heating process:

physics calorimetry heat exchange calorimeter measuring thermal quantities by the method of mixtures

  1. the temperature of $A$ rises twice as much as $B$

  2. the temperature of $A$ rises four times as much as $B$

  3. the temperature of $B$ rises twice as much as $A$

  4. the temperature of $B$ rises four times as much as $A$

  5. the temperature of both $B$ and $A$ rise the same amount

Show answer
Correct Option: C
Explanation:

Let specific heat of substance $A$  is $2c$ and specific heat of substance $B$ is $c$ , 

we have ,  heat given  $Q=mc\Delta t$, where  $\Delta t $ denotes change in temperature , 
so , for substance $A$, $Q=m\times 2c\Delta t _{A}$ 
or $\Delta t _{A}=Q/2mc$ .........eq1
For substance $B$ ,   $Q=mc\Delta t _{B}$   
$\Delta t _{B}=Q/mc$ ...................eq2
by eq1 and eq2,
$2\Delta t _{A}=\Delta t _{B}$     

To measure the specific heat of copper, an experiment is performed in the lab. A piece of copper is heated in an oven then dropped into a beaker of water. To calculate the specific heat of copper, the experimenter must know or measure the value of all of the quantities below EXCEPT the

physics calorimetry heat exchange calorimeter measuring thermal quantities by the method of mixtures

  1. Original temperatures of the copper and the water

  2. Mass of the water

  3. Final (equilibrium) temperature of the copper and the water

  4. Time taken to achieve equilibrium after the copper is dropped into the water

  5. Specific heat of the water

Show answer
Correct Option: B
Explanation:

Specific of  a substance  is given by: $\Delta Q$= $mc\Delta T$

where, $\Delta Q=$ heat given to substance
                $m=$ mass  of  the substance
             $\Delta T=$ increase in temperature of substance (for that we require initial and final temperature of substance)
If the substance is copper in this experiment the experimenter requires the mass of copper piece not the mass of water because water is just dropping the temperature of copper  piece not more  than this.        

An aluminium block of 2m mass and an iron block of m mass,each absorbs the same amount of heat, and both blocks remain solid. If the specific heat of aluminium is twice the specific heat of iron, then find out the correct statement?

physics calorimetry heat exchange calorimeter measuring thermal quantities by the method of mixtures

  1. The increase in temperature of the aluminum block is twice the increase in temperature of the iron block

  2. The increase in temperature of the aluminum block is four times the increase in temperature of the iron block

  3. The increase in temperature of the aluminum block is the same as increase in temperature of the iron block

  4. The increase in temperature of the iron block is twice the increase in temperature of the aluminum block

  5. The increase in temperature of the iron block is four times the increase in temperature of the aluminum block

Show answer
Correct Option: E
Explanation:

The heat required to rise the temperature of body of mass $m$ of specific heat $s$ by $\Delta T=H=ms\Delta T$

Thus for same amount of heat, the rise in temperature ration of aluminium and iron is $\dfrac{\Delta T _{Al}}{\Delta T _{Fe}}=\dfrac{m _{iron}s _{iron}}{m _{aluminium}s _{aluminium}}=\dfrac{1}{4}$
Thus the rise in temperature of the iron block is four times the increase in temperature of the aluminum block

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.

physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

  1. 200 J

  2. 400 J

  3. 600 J

  4. 800 J

  5. 1,000 J

Show answer
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 } $)

physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

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

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

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

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

Show answer
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 $\Delta T=4^oC$

$5gm$ of steam at $100^oC$ is passed into calorimeter containing liquid , Temperature of liquid rises from $32^oC$ to $40^oC$. Then water equivalent of calorimeter and content is 

physics calorimetry heat exchange calorimeter measuring thermal quantities by the method of mixtures

  1. $40$ gram

  2. $375$ gram

  3. $300$ gram

  4. $160$ gram

Show answer
Correct Option: B
Explanation:

Latent heat of vaporization of water = $2260kJ/Kg$

The specific heat capacity of water = $4185.5 J/Kg$

Heat lost by steam = heat gained by the water and calorimeter

Formula :

Heat gained by water = mcФ

m = mass of water

c = specific heat capacity of water.

$Ф = Change in temperature.  = 40 - 32 = 8$

Heat lost by steam = mLv + mcФ

Lv = latent heat of vaporisation

m = mass of steam.

$Ф = 100 - 40 = 60$

Doing the substitution :

$2260000 \times 0.005 + 60 \times 0.005 \times 4185.5 = m \times 4185.5 \times 8$

$12555.65 = 33484m$

$m = \dfrac {12555.65}  {33484} = 0.37497 kg$

$= 0.37497 \times 1000 = 374.97 kg$

$= 374.97kg$