Tag: measuring thermal quantities by the method of mixtures

Questions Related to measuring thermal quantities by the method of mixtures

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

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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