Tag: measuring thermal quantities by the method of mixtures

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

A piece of copper weighing 500 g is heated to $100^oC$ and dropped into 200g of water at $25^oC$. Find the temperature of the mixture. The specific heat of Cu is $0.42 J g^{-1} {\;}^oC^{-1}$.

  1. <span>$30^oC$</span>

  2. <span>$40^oC$</span>

  3. <span>$50^oC$</span>

  4. <span>$60^oC$</span>
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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 $40^oC$.

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

$10\ kg$ of hot water in a bucket at $70^oC$ is cooled for taking a bath adding to it $20\ kg$ water at $20^oC$. What is the temperature of the mixture? (Neglect the thermal capacity of the bucket)

  1. $30.67^oC$

  2. $36.67^oC$

  3. $60.67^oC$

  4. $46.67^oC$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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 $30T _f=1100 \therefore T _f=36.67^oC$.

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

What is the final temperature of the mixture of 300 g of water at $25^oC$ added to 100 of ice at $0^oC$.

  1. <span>$0^oC$</span>

  2. <span>$1^oC$</span>

  3. <span>$2^oC$</span>

  4. <span>$3^oC$</span>
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

$\therefore 300 (25-\theta)=100\times 80+100\times 0.5\theta$
$\therefore \theta=-\frac {5}{3.5}$
Since $\theta$ is negative
Heat lost is utilised to melt only same part of ice. Hence equilibrium temperature is $0^oC$.

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

500 g of water at $100^oC$ is mixed with 300 g at $30^oC$. Find the temperature of the mixture. Specific heat of water $=4.2 J g^{-1} {\;}^oC^{-1}$.

  1. $73.8^oC$

  2. $53.8^oC$

  3. $40^oC$

  4. $60^oC$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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 $73.8^oC$.

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

A calorimeter contains $70.2 \,g$ of water at $15.3^o C$. If $143.7 \,g$ of water at $36.5^o C$ in mixed it with the common temperature is $28.7^o C$. The water equivalent of the calorimeter is:

  1. $15.6 \,g$

  2. $9.4 \,g$

  3. $6.3 \,g$

  4. $13.4 \,g$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

$5$g of copper was heated from $20^{\circ}$ to $80^{\circ}$. How much energy was used to heat Cu? (Specific heat capacity of Cu is $0.092 cal/g ^{\circ}C$).

  1. $27.6$ cal

  2. $50$ cal

  3. $35$ cal

  4. $25.7$ cal
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given,

Mass, $m=5\,g$

Specific heat capacity, $C=0.092\,cal/g \,^0C$

Change in temperature, $\Delta T=80\,^0C -20^0C=600^0C$

Heat required, $Q=?$

We have the equation,

$Q=m\times C\times \Delta T$

Then,

$Q=5\times 0.092\times 60=27.6\,cal$
Multiple choice physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

A calorimeter constains 10 g of water at ${ 20 }^{ \circ  }$ C. The temperature falls to ${ 15 }^{ \circ  }$ C in 10 min. When calorimeter contains 20 g of water at ${ 20 }^{ \circ  }$ C, it takes 15 min for the temperature to become ${ 15 }^{ \circ  }$ C. The water equivalent of the calorimeter is

  1. 5 g

  2. 10 g

  3. 25 g

  4. 50 g
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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.

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

What is the principle of the method of a mixture? Name the law on which this principle is based.

  1. &nbsp;Newtons law of cooling&nbsp;

  2. none

  3. <span>principle of calorimetry&nbsp;</span>

  4. <span>principle of heat transfer</span>
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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.

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

In a calorimeter of water equivalent  $20 { g },$  water of mass  $1.1 { kg }$  is taken at  $288{ K }$  temperature. If steam at temperature  $373 { K }$  is passed through it and temperature of water increases by  $6.5 ^ { \circ } { C }$  then the mass of steam condensed is

  1. $17.5{ g }$

  2. $11.7{ g }$

  3. $15.7{ g }$

  4. $18.2{ g }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

2000 J of energy is needed to heat 1 kg of paraffin through $1^{\circ}C$. So How much energy is needed to heat 10 kg of paraffin through $2^{\circ}C$ ?

  1. 4000 J

  2. 10,000 J

  3. 20,000 J

  4. 40,000 J
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

we get value of c
Hence for $10 kg$ through $2^oC$,
$W= (10000)(2)(2)= 40000 J$