Tag: measurement and effects of heat

Questions Related to measurement and effects of heat

500 g of hot water at $60^oC$ is kept in the open till its temperature falls to $40^oC$. Calculate the heat energy lost to the surroundings by the water. (Specific heat of water $=4200 J kg^{-1} {\;}^oC^{-1})$

  1. 2400 J

  2. 5000 J

  3. 40000 J

  4. 42000 J


Correct Option: D
Explanation:

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

How much amount of heat is required to raise the temperature of 100 g of water from $30 ^oC$ to $100 ^oC$? The specific heat of water $=4.2 J g^{-1} {\;}^oC^{-1}$.

  1. 25.5 kJ

  2. 29.4 kJ

  3. 30 kJ

  4. 40 kJ


Correct Option: B
Explanation:

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 $Q=m . C. \Delta T$
$=100 g\times 4.2 J g^{-1} {\;}^oC^{-1}\times 70 ^oC$
$=29400 J=29.4 kJ$

What quantity of heat would be given out by 200 gm of copper in cooling from $80^oC$ to $20^oC$ (Specific heat of copper $=0.09 cal g^{-1} {\;}^oC^{-1})$?

  1. 1080 cal

  2. 1000 cal

  3. 1500 cal

  4. 1100 cal


Correct Option: A
Explanation:

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

If $20 \ g$ of ice at $0^\circ C$ is mixed with $10 \ g$ of water at $40^\circ C$, the final mass of water in the mixture is:

  1. 10 g

  2. 25 g

  3. 18 g

  4. 20 g


Correct Option: A

A liquid P if specific heat capacity $2400 J kg^{-1} K^{-1}$ and at $70^oC$ is mixed with another liquid R of specific heat capacity $1000 J kg^{-1} K^{-1}$ at $30^oC$. After mixing, the final temperature of the mixture is $40^oC$. Find the ratio of the mass of the liquids mixed?

  1. 4 : 5

  2. 8 : 5

  3. 40 : 5

  4. 48 : 5


Correct Option: D
Explanation:

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.
$\therefore m _PC _P(70-40)=m _RC _R(40-30)$
$m _P(2400)(40)=m _R(1000)(10)$
$\frac {m _P}{m _R}=\frac {2400\times 40}{1000\times 10}=\frac {48}{5}$
$\frac {m _P}{m _R}=48 : 5$

Calculate the amount of heat required to convert 5 kg of ice to $0^oC$ to vapour at $100^oC$.

  1. $1.5\times 10^7 J$

  2. $2.5\times 10^7 J$

  3. $3.5\times 10^7 J$

  4. $4.5\times 10^7 J$


Correct Option: A
Explanation:

$m=5 kg, L _f=3.36\times 10^5 J kg^{-1}$
$L _V=2.25\times 10^6 J kg^{-1}$
$C=4200 J kg^{-1} {\;}^oC^{-1}$ 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 $100^oC$
$=mL _f+ms(100-0)+mL$
$=5\times 3.36\times 10^5+5\times 4200\times (100-0)+5\times 2.25\times 10^6$
$=16.8\times 10^5+21\times 10^5+112.5\times 10^5$
$=1.5\times 10^7J$

Utensils used for efficient cooking should have

  1. Large heat capacity

  2. Small heat capacity

  3. Medium heat capacity

  4. Any heat capacity


Correct Option: A

A copper calorimeter of a mass $300\ g$ contains $500\ g$ of water at a temperature of $20^\circ C$. A $500\ g$ of copper block at $100^\circ C$ is dropped into the calorimeter. If the resultant temperature is $25^\circ C$, then fond the specific heat of copper in $JKg^{-1} K^{-1}$.

  1. $190$

  2. $290$

  3. $390$

  4. $490$


Correct Option: C

Calories is defined as the amount of heat required to rise temperature of $1\ g$ of water by $1^{o}C$ and it is defined under which of the following conditions.

  1. From $14.5^{o}C$ to $15.5^{o}C$ at $760\ mm$ of $Hg$

  2. From $98.5^{o}C$ to $99.5^{o}C$ at $760\ mm$ of $Hg$

  3. From $13.5^{o}C$ to $14.5^{o}C$ at $76\ mm$ of $Hg$

  4. From $3.5^{o}C$ to $4.5^{o}C$ at $76\ mm$ of $Hg$


Correct Option: A

 Which of the following material is used to make calorimeter?

  1. Glass

  2. Ebonite

  3. Metal

  4. Superconductor


Correct Option: C