Tag: measuring thermal quantities by the method of mixtures

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Multiple choice physics calorimetry heat exchange calorimeter 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.

  1. $24.44^{\circ} C$

  2. $46.67^{\circ} C$

  3. $44.44^{\circ} C$

  4. $54.44^{\circ} C$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $\theta _0=46.67^o$
Multiple choice physics calorimetry heat exchange calorimeter measuring thermal quantities by the method of mixtures

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:

  1. $18.2^ C$

  2. $22^ C$

  3. $20.3^ C$

  4. $24.2^ C$
Reveal answer Fill a bubble to check yourself
C Correct answer
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
$ms _1(t-12)=ms _3(28-t)$
$\dfrac{15}{16}s _2(t-12)=s _3(28-t)$
$15t-180=448-16t$
$31t=448+180=628$
$t=\dfrac{628}{3}=20.3^o C$

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

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:

  1. 1 litre

  2. 2 litre

  3. 3 litre

  4. 4 litre
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $V _p$ =  $\displaystyle \frac{0.32}{80}$
$= \dfrac{1}{250} m^3=\dfrac{1000}{250} litre = 4 litre$.

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

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

  1. $1 : 3$

  2. $2 : 3$

  3. $1 : 1$

  4. $3 : 4$
Reveal answer Fill a bubble to check yourself
C Correct answer
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