Tag: calorimetry

Questions Related to calorimetry

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

$5gm$ 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 :

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

$12555.65 = 33484m$

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

$= 0.37497 \times 1000 = 374.97 kg$

$= 374.97kg$

1 kg of water at $20^{\circ}C$ is, mixed with 800 g of water at $80^{\circ}C$. Assuming that no heat is lost to the surroundings. Calculate the final temperature of the mixture.

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

  1. $24.44^{\circ} C$

  2. $46.67^{\circ} C$

  3. $44.44^{\circ} C$

  4. $54.44^{\circ} C$

Show answer
Correct Option: B
Explanation:

$mc\theta _1 + mc\theta _2 = mc\theta _0$
since c is constant, assuming water is heated from $0^o C$
(1000)(20) + (800)(80)= (1800)$\theta _0$
20000+64000= 84000=1800$\theta _0$
$\theta _0= 46.67^o C$
hence the finale temperature in $\theta _0=46.67^o$

The temperature of equal masses of three different liquids A, B, and C are $12^o C$,$19^o C$ and $28^o C$ respectively. The temperature when A and B are mixed is $16^oC$ and When B and C are mixed is $23^o C$. The temperature when A and C are mixed is:

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

  1. $18.2^ C$

  2. $22^ C$

  3. $20.3^ C$

  4. $24.2^ C$

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Correct Option: C
Explanation:

let $m _1=m _2=m _3=m$
Let $s _1,s _2,s _3$ be the respective specific heats of the three liquids.
When A and B are mixed, temperature of mixed = $16^o C$
A heat gained by A = heat lost by B
$\therefore ms _1(16-12)=ms _2(19-16)$
$4s _1=3s _2$.......(i)
When B and C are mixed , temperature of mixture$=23^o C$.
As heat gained by B = heat lost by C,
$ms _2(23-19)=ms _3(28-23)$
$\therefore 4s _2=5s _3$......(ii)
From (i) and (ii) , $=\dfrac{3}{4}s _2=\dfrac{15}{16}s _3$
When A and C are mixed, suppose temperature of mixture$=t$
Heat gained by A = Heat lost by C
$ms _1(t-12)=ms _3(28-t)$
$\dfrac{15}{16}s _2(t-12)=s _3(28-t)$
$15t-180=448-16t$
$31t=448+180=628$
$t=\dfrac{628}{3}=20.3^o C$

An adulterated sample of milk has a density, 1032 kg m$^{-3}$, while pure milk has a density of 1080 kg m$^{-3}$. Then the volume of pure milk in a sampled of 10 litres of adulterated milk is:

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

  1. 1 litre

  2. 2 litre

  3. 3 litre

  4. 4 litre

Show answer
Correct Option: D
Explanation:

Mass of adulterated milk
$M _A = 1032 \times (10 \times 10^{-3}$) kg
or $M _A = 10.32 kg$      ($\because 1 litre = 10^{-3} m^3$)
$\therefore$ Mass of pure milk $M _p= 1080 \times V _p$ 

$\therefore$ Mass of water added = $\rho _wV _w= M _A - M _p$
$\therefore$ 10$^3 \times$ (volume of water added)= $M _A - M _p$
$\therefore10^3\times(10 \times 10^{-3}- V _p) = 10.32 - 1080 V _p$
or 80 $V _p = 0.32$
or $V _p$ =  $\displaystyle \frac{0.32}{80}$
$= \dfrac{1}{250} m^3=\dfrac{1000}{250} litre = 4 litre$.

An experiment requires a gas with $\gamma = 1.50$. This can be achieved by mixing together monatomic and rigid diatomic ideal gases. The ratio of moles of the monatomic to diatomic gas in the mixture is

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

  1. $1 : 3$

  2. $2 : 3$

  3. $1 : 1$

  4. $3 : 4$

Show answer
Correct Option: C
Explanation:

One mole of an ideal monoatomic gas is is C$ _{v}$ = $\dfrac{3}{2}$R and C$ _{p}$ = $\dfrac{5}{2}$R


i.e $\gamma$ = 1.66 for monoatomic gas

For One mole of an ideal dioatomic gas,
$\gamma$ = 1.4 for air which is pre dominantly a  diatomic gas
If we take 1 mole monoatomic and 1 mole of diatomic gas in a mixture then we get the following result;

$\gamma$ = $\dfrac{n1\gamma + n2\gamma}{n1 + n2}$ 

Now since we have taken the no. of moles of monoatomic as well as diatomic as 1, therefore
$\gamma$ = $\dfrac{y1 + y2}{2}$ where $\gamma$1 and $\gamma$2 are the values of $\dfrac{C _p}{C _v}$ for individual gases.

Substuting the values of C$ _p$ and C$ _v$ i.e $\gamma$1 = 1.6 and $\gamma$2 = 1.4 we get
$\gamma$ = 1.53 which is approximately equal to 1.50 which is given.
Hence by taking 1 mole og monoatomic and 1 mole of diatomic mixture we got $\gamma$ as 1.50
Hence the ratio of moles of monoatomic to diatomic gas in the mixture is 1:1