Tag: calorimetry

Quizzes (3) Questions (19)

Questions Related to calorimetry

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 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 $2^oC$,
$W= (10000)(2)(2)= 40000 J$

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

Which of the following properties must be known in order to calculate the amount of heat needed to melt 1.0kg of ice at $0^oC$? 
I. The specific heat of water 
II. The latent heat of fusion for water 
III. The density of water.

  1. I only

  2. I and II only

  3. I, II, and III

  4. II only

  5. I and III only
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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 ,

            $Q=mL$ ,  where $m=$ mass of substance , $L=$ latent heat
 here we have $m=1.0kg$ but we don't have value of $L$ (latent heat of fusion for water) so it is required .
    Density and specific heat of water are not required here , as it is clear from formula mentioned above .

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

One calorie is defined as the heat required to raise the temperature of $1$ gm of water by $1^o$C in a certain interval of temperature and at certain pressure. The temperature interval and pressure is?

  1. $13.5^o$ C to $14.5^o$ C & $76$ mm of Hg

  2. $6.5^o$ C to $7.5^o$ C & $76$ mm of Hg

  3. $14.5^o$ C to $15.5^o$ C & $760$ mm of Hg

  4. $98.5^o$ C to $99.5^o$ C & $760$ mm of Hg
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

in $760\ mm$ of $Hg$.

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

A copper ball of mass $100gm$ is at a temperature $T$. It is dropped in a copper calorimeter of mass $100gm$, filled with $170gm$ of water at room temperature. Subsequently the temperature of the system is found to b4 ${75}^{o}$. $T$ is given by then (Given: room temperature $={30}^{o}C$, specific heat of copper $=0.1cal/gm _{  }^{ o }{ C }\quad $)

  1. ${ 825 }^{ o }C$

  2. ${ 800 }^{ o }C$

  3. ${ 885 }^{ o }C$

  4. ${ 1250 }^{ o }C$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Final temperature of celomiter and its constant is given as
$To=75^o C$
$\Rightarrow \ 100\times 0.1\times (75-T)+100\times 0.1(75-30)+1.70 \times 1\times (75.32)$
$\therefore \ T=885^oC$
Multiple choice physics calorimetry heat exchange calorimeter measuring thermal quantities by the method of mixtures

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

  1. $17.5g$

  2. $11.7g$

  3. $15.7g$

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

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.

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

Steam at $100^oC$ is passed into $2.0$kg of water contained in a calorimeter of water equivalent $0.02$kg at $15^oC$ till the temperature of the calorimeter and its content rise to $90^oC$. The mass of steam condensed in kg is

  1. $0.301$

  2. $0.280$

  3. $0.60$

  4. $0.02$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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.

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

A copper calorimeter contains $100 g$ of water at $16^o C$. When $15 g$ of ice is added to it, the resultant temperature of the mixture is $4^o C$. Water equivalent of the calorimeter is

  1. $8 g$

  2. $12 g$

  3. $6 g$

  4. None
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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.

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

$50  g$ of ice at 0 C is mixed with $50  g$ of water at 20 C.The resultant temperature of the mixture would be

  1. 10 C

  2. 0 C

  3. -10 C

  4. -35 C
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$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.