Measuring thermal quantities by the method of mixtures - class-XI

Given $m=150$
$g=\frac {150}{1000}=0.15 kg$
Specific heat of iron
$C=480 J kg^{-1} {;}^oC^{-1})$
$\Delta =(25-20)^oC=5^oC$
$Q=m\times C\times \Delta T$
$=0.15 kg\times 480 J kg^{-1} {;}^oC^{-1}\times 5^oC$
$=360 J$.

$m=400 g=0.4 kg$
$C=390 J kg^{-1} {;}^oC^{-1}$
$T _1=100 ^oC, T _2=30^oC$
$\Delta T=70^oC$
$\therefore$ Heat lost $=mC(T _1-T _2)$
$=0.4\times 390\times 70 J=10,920 J$

$Q=1680 J$
$m=1000 g$ or 1 kg

Principle of calorimetry states that heat lost by a hotter body = heat gained by a colder body, therefore calorimetry is a process in which heat energy of both hot and cold body equalizes.

$m=100 g, C=1 cal g^{-1} {;}^oC^{-1}$
$\Delta T=(95-5)^oC=90^oC$
$Q=mC\Delta T$
$=100 g\times 1 cal g^{-1} {;}^oC^{-1}\times 90^oC$
$=100\times 1\times 90 cal$
$=9000 cal=9 k cal$.

Hot water, $m _h=5 kg, T _h=80^oC$
Cold water, $m _c=15 kg, T _c=20^oC$
$T=?$
If the temperature of mixture is T
Heat lost by hot water
$5\times C\times (80-T)$
Heat gained by cold water
$15\times C\times (T-20)$
According to the principle of calorimetry, Heat lost $=$ Heat gained
$5\times C\times (80-T)=15\times C\times (T-20)$
$\therefore 80-T=3(T-20)$

Heat lost by 5 g of water at $30^oC$ to water at $0^oC$.
$Q _L=5\times 1\times 30=150 cal$
Heat required by 5 g of ice at $-20^oC$
$(Q _1)=5\times 0.5\times (2.0)=50 cal$
Heat required by 5 g of ice at $0^oC$  into water at $0^oC$.
$(Q _2)=5\times 80=400 cal$
$Q _L < (Q _1+Q _2)$
$\therefore$ The final temperature of the mixtuure is $0^oC$.

$C=0.5 cal g^{-1} {;}^oC^{-1}$

Heat lost by 0.05 kg of steam at $100^oC$ to water at $0^oC$
$Q _L=(50\times 540)+(50\times 1\times 100)$
$=27000+5000=32000 cal$
Heat required by 400 g of ice at 253 K $(-20^oC)$ to convert into water at 273 K $10^oC) Q _1=(400\times 0.5\times 20)+(400\times 80)$
$=(4000+32000)=36000 cal$
$\therefore Q _L < Q _1$, only part of ice melts and final temperature remains $0^oC$ or 273 K.

Given, mass of copper, $m _1=500 g$
Mass of water, $m _2=200 g$
Initial temperature of copper, $t _1=100^oC$
Initial temperature of water, $t _2=25^oC$
Sp. heat of copper, $C _1=0.42 J g^{-1} {;}^oC^{-1}$
Sp. heat of water, $C _2=4.2 J g^{-1} {;}^oC^{-1}$
Final temperature of the mixture $=t^oC$
Then,
Heat lost by the copper piece
$=m _1C _1(t _1-t)$
Heat gained by water $=m _2C _2(t-t _2)$
We know, Heat lost $=$ Heat gained
$\Rightarrow m _1C _1(t _1-t)=m _2C _2(t-t _2)$
$\Rightarrow 500\times 0.42\times (100-t)$
$=200\times 4.2\times (t-25)$
$\Rightarrow (100-t)=\frac {200\times 4.2}{500\times 0.42}\times (t-25)$
$=4(t-25)$
This given, $5t=200$
$\Rightarrow t=\frac {200}{5}^oC=40^oC$
Thus, the final temperature of the mixture is $40^oC$.

m (hot water) $=10 kg$,
T (hot water) $=70^oC$
m (cold water) $=20 kg$
T (cold water) $=20^oC, T (final)=?$
Using the formula $Q=mC\Delta t$
We get heat lost by hot water
$=10\times C\times (70-T _f)$
Where $T _f$ is the final temperature
Heat gained by cold water
$=20\times C\times (T _f-20)$
Using the principle of calorimetry
Heat lost $=$ Heat gained
We get $10\times C\times (70-T _f)$
$=20\times C\times (T _f-20)$
$\therefore 700-10T _f=20T _f-400$
or $30T _f=1100 \therefore T _f=36.67^oC$.

Heat lost by a hot body $=$ Heat gained by a

Mass of hot water, $m _1=500 g$
Mass of cold water, $m _2=300 g$
Temp. of hot water, $t _1=100^oC$
Temp. of cold water, $t _2=30^oC$
Sp. heat of water, $C=4.2 J g^{-1} {;}^oC^{-1}$
Let temp. of mixture be $t^oC$. Then, Heat gained by cold water
$=m _2\times C\times (t-t _2)$
According to the principle of calorimetry, Heat lost $=$ Heat gained
$500\times 4.2\times (100-t)$
$=300\times 4.2\times (t-30)$
$\Rightarrow 5(100-t)=3(t-30)$
$\Rightarrow -3t-5t=-90-500$
$\Rightarrow -8t=-590$
$\Rightarrow t=\frac {590}{8}=73.8^oC$
So, the final temperature of the mixture is $73.8^oC$.

Assume water equilient is $= ngm$
So, $(70.2) \times 1 \times (28.7 - 15.3) + n (28.7 - 15.3) = 143.7 \times 1 (36.5 - 28.7)$
$n (13.4) = 180.18$
$n = 13.4 ,g$

Using Newton's law of cooling, the rate of cooling is proportional to the temperature difference. By setting up two equations for the two cases (10g and 20g of water), the water equivalent of the calorimeter can be calculated as 25g.

A principle of calorimetry states that if there is no loss of heat in surrounding the total heat loss by hot body equals to total heat gained by a cold body.
i.e. heat loss = heat gained
It is based on Newton's Law of Cooling, which states that when a liquid is heated of higher temperature and placed to cool. Then the rate of heat lost by a temperature of the liquid is directly proportional to the difference in temperature of the surrounding.

The heat lost by the steam (condensing and cooling) equals the heat gained by the water and the calorimeter. Using the formula mL + mcdT = (M_waterc + W_cal)*dT, solving for m yields approximately 11.7g.

$W= mc\theta$
$2000= (1000)c(1)$
$c= 2$ $J/g^oC$

Energy for ice= mL=440(336)= 147840 J
Energy in water= mc$\theta$= 540(80)(4.2)= 181440J

Heat required $=mS\triangle T \ =2000\times 1\times (50-10) \ =80000cal$

Principle of calorimetry states that heat lost by a hotter body = heat gained by a colder body, therefore in calorimetry, total heat energy of the system remains constant, which is the law of conservation of energy . 

We know that, $heat=mass\times specific ;heat\times change;in;temperature \ \Rightarrow 60=15\times 1\times \triangle T \ \Rightarrow \triangle T=4^oC$

Calorimetry is the branch of physics which deals with the measurement of heat energy. Calorimetry is one of the methods for the determination of specific heats or latent heats of the substances.

The latent heat is the heat required to change the state of unit mass of substance ,  therefore heat required to change the mass $m$ of substance  is given by ,

When heat is added to ice at $0^{\circ}C$, the temperature of the ice does not change. However the heat goes into the latent heat of fusion of ice, This is a kind of potential energy that water owns in form of latent heat.

Given ,  $m=400g ,  \theta _{1}=100^{0}C , \theta _{2}=?$ , specific heat of  vegetable oil $c=1.98J/g-^{o}C , Q=47376J$

Given ,  $m=150g ,  \theta _{1}=20^{0}C , \theta _{2}=25^{0}C$

Given ,  $m=100g ,  \theta _{1}=5^{0}C , \theta _{2}=95^{0}C$

Given ,  $m=200g , Q=2000cal ,  \Delta\theta=? , $ , specific heat of water $c=1cal/g-^{o}C$

Amount of heat required to convert unit mass of ice into water is called latent heat ($L$) of fusion of ice  i.e.

We have ,  $L _{ice}=3.34\times10^{5}J/kg , m=1.5kg$

One calories is defined as the amount of heat required to raise the temp of $1\ gm$ of water from $14.5^oC$ to $15.5^oC$

$m=400 g, C=1.98 Jg^{-1} {;}^oC^{-1}$
$T _1=100^oC, T _2=?$
Fall in temperature
$\Delta T=(100-x)$
Heat energy given out by oil
$=47376 J$
According to formula $Q=m.C.\Delta T$
$\Rightarrow 47376=400\times 1.98 (100-x)$
$\Rightarrow 100-x=\frac {47376}{400\times 1.98}=59.8$
$\Rightarrow x=100-59.8=40.2^oC$
$\therefore$ Final temperature of oil $=40.2^oC$.

Quantity of heat supplied
$Q=2000 cal$
Mass of water, $m=200 g$
Specific heat of water
$C=1 cal g^{-1} {;}^oC^{-1}$
Rise in temperature $=?$
From relation, $Q=m C\Delta T$
$\Rightarrow \Delta T=\frac {Q}{m.C}$
$=\frac {2000 cal}{200 g\times 1cal g^{-1} {;}^oC^{-1}}=10^oC$
So, the temperature of water rises by $10^oC$.

$m=500, g=0.5 kg$
$\Delta T=60-40=20^oC$
$C=4200 J kg^{-1} {;}^oC^{-1} Q=?$
Using the formula, $Q=m C \Delta T$
$=0.5\times 4200\times 20=42000 J$
$\therefore \text {Heat lost}=42,000 J$

Mass of water, $m=100 g$
Rise in temperature $(\Delta T)$
$=(100^oC-30^oC)=70 ^oC$
Specific heat of water
$C=4.2 J g^{-1} {;}^oC^{-1}$
Then $Q=m . C. \Delta T$
$=100 g\times 4.2 J g^{-1} {;}^oC^{-1}\times 70 ^oC$
$=29400 J=29.4 kJ$

$m=200 g, C _{copper}=0.09 cal g^{-1} {;}^oC^{-1}$
$\Delta T=T _1-T _2=80-20=60^oC$
$\therefore Q=mC \Delta T=200\times 0.09\times 60$
$=1080 cal$.

Let mass of the liquids P and R be $m _P$ and $m _R$ respectively.
According to calorimeter,
Heat lost by liquid P $=$ Heat gained by liquid R.
$\therefore m _PC _P(70-40)=m _RC _R(40-30)$
$m _P(2400)(40)=m _R(1000)(10)$
$\frac {m _P}{m _R}=\frac {2400\times 40}{1000\times 10}=\frac {48}{5}$
$\frac {m _P}{m _R}=48 : 5$

$m=5 kg, L _f=3.36\times 10^5 J kg^{-1}$
$L _V=2.25\times 10^6 J kg^{-1}$
$C=4200 J kg^{-1} {;}^oC^{-1}$ Then $Q=?$
$Q=$ Heat required to convert ice at $^oC$ to water $0^oC +$ Heat required to convert water at $100^oC$ to vapour at $100^oC$
$=mL _f+ms(100-0)+mL$
$=5\times 3.36\times 10^5+5\times 4200\times (100-0)+5\times 2.25\times 10^6$
$=16.8\times 10^5+21\times 10^5+112.5\times 10^5$
$=1.5\times 10^7J$

This is identical to 528091. The heat gained by the water (1.1kg) and calorimeter (20g) for a 6.5C rise equals the heat released by condensed steam. The calculation confirms 11.7g.

Heat gained by water (2kg) and calorimeter (0.02kg) = (2.02kg * 1000g/kg * 1 cal/gC * (90-15)C) = 151500 cal. Heat lost by steam = m * (540 + (100-90)) = m * 550. m = 151500 / 550 = 275.45g = 0.275kg, which rounds to 0.280kg.

The calorie is defined as the amount of heat required to raise the temperature of 1g of water from 14.5C to 15.5C at standard atmospheric pressure (760 mm Hg).

Heat lost by water and calorimeter = (100 + W) * (16-4) = 12(100+W). Heat gained by ice = 15 * 80 + 15 * (4-0) = 1200 + 60 = 1260. 12(100+W) = 1260 => 100+W = 105 => W = 5g. Since 5g is not an option, 'None' is the correct choice.

Calorimeters are made of metals like copper or aluminum because they have high thermal conductivity, ensuring the contents reach thermal equilibrium quickly.

Water equivalent = mass * specific heat. 400g * 0.1 cal/gC = 40g.

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 $98.6 ^{\circ} F$

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 $0^0C$ 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.

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

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

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

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

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

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

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

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 :

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

$12555.65 = 33484m$

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

$= 0.37497 \times 1000 = 374.97 kg$

$= 374.97kg$

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
$ms _1(t-12)=ms _3(28-t)$
$\dfrac{15}{16}s _2(t-12)=s _3(28-t)$
$15t-180=448-16t$
$31t=448+180=628$
$t=\dfrac{628}{3}=20.3^o C$