# Calorimetry

## Trending Questions

**Q.**20 gm ice at −10 ∘C is mixed with m gm steam at 100 ∘C. The minimum value of m so that finally all ice and steam converts into water at 0 ∘C is

(Use: specific heat and latent heat as Cice=0.5 cal/gm ∘C, Cwater=1 cal/gm ∘C, Lfusion =80 cal/gm and Lvapor =540 cal/gm )

- 18527 gm
- 13517 gm
- 8532 gm
- 11317 gm

**Q.**When m grams of water at 10∘C is mixed with m grams of ice at 0∘C, which of the following statements are false?

- The temperature of the system will be given by the equation m×80+m×1×(T−0)=m×1×(10−T)
- Whole of ice will melt and temperature will be more that 0∘C but lesser than 10∘C
- Whole of ice will melt and temperature will be 0∘C
- Whole of ice will not melt and temperature will be 0∘C

**Q.**Figure shows the temperature variation when heat is added continuously to a specimen of ice (10 g) at −40∘C at constant rate. (Specific heat of ice =0.53 cal/g∘C and Lice=80 cal/g, Lwater=540 cal/g)

Column - IColumn - II(A)Value of Q1(in cal)(P)800(B)Value of Q2(in cal)(Q)1000(C)Value of Q3(in cal)(R)5400(D)Value of Q4(in cal)(S)212(T)900

- A→S;B→P;C→Q;D→T
- A→P;B→S;C→Q;D→R
- A→P;B→S;C→R;D→Q
- A→S;B→P;C→Q;D→R

**Q.**The water equivalent of a copper calorimeter is 4.5 g. If the specific heat of copper is 0.09 cal g−1C−1, then

- Mass of the calorimeter is 0.5 kg.
- Thermal capacity of the calorimeter is 4.5 cal∘C−1.
- Heat required to raise the temperature of the calorimeter by 8∘C will be 36 cal
- Heat required to melt 15 gm of ice placed in the calorimeter will be 1200 cal

**Q.**The water equivalent of 20 g of aluminium (specific heat 0.2 cal g−1 ∘C−1) , is

- 40 g
- 4 g
- 8 g
- 160 g

**Q.**The amount of heat required to convert 5 g of ice at 0∘C to 5 g of steam at 100∘C is -

[Latent heat of vaporization and fusion are Lv=540 cal g−1 and Lf=80 cal g−1]

- 3100 cal
- 3200 cal
- 3600 cal
- 4200 cal

**Q.**

Give an example, where high specific heat capacity of water is used as a heat reservoir.

**Q.**Four cubes of ice at −10∘C, each of one gram, is taken out from a refrigerator and dropped into 150 g of water at 20∘C. Find the temperature of the water when thermal equilibrium is attained. Assume that no heat is lost to the outside and the water equivalent of the container is 46 g. (Specific heat capacity of water =1 cal/g∘C, specific heat capacity of ice =0.5 cal/g∘C, latent heat of fusion of ice =80 cal/g)

- 9.1∘C
- 17.9∘C
- 27.2∘C
- 37.4∘C

**Q.**A 5g piece of ice at −20∘C is put into 10g of water at 30∘C. Assuming that heat is enchanged only between the ice and the water, find the final temperature of the mixture. Specific heat capacity of ice =2100Jkg−1∘C−1. Specific heat capacity of water =4200Jkg−1∘C−1 and latent heat of fusion of ice 3.36×105 J kg−1.

- 0∘ C
- 10∘ C
- 20∘ C
- 30∘ C

**Q.**A body of mass 5 kg falls from a height of 30 metres. If all of its mechanical energy is converted into heat, then heat produced will be

[Take g=9.8 m/s2 and J=4.2 J/cal]

- 350 J
- 350 cal
- 375 J
- 375 cal

**Q.**100 g of ice (latent heat 80 cal g−1) at 0∘C is mixed with 100 g of water (specific heat 1 cal g−1 ∘C−1) at 80∘C. The final temperature of the mixture will be:

- 0∘C
- 40∘C
- 80∘C
- <0∘C

**Q.**On which of the following does the heat capacity of a body depend?

- The heat input
- The mass of the body
- The initial temperature
- The material it is made up of
- Volume of the system under consideration

**Q.**

A lead bullet penetrates into a solid object and melts. Assuming that 50% of its kinetic energy was used to heat it, calculate the initial speed of the bullet. The initial temperature of the bullet is 27∘C and its melting point is 327∘C. Latent heat of fusion of lead = 2.5 × 104 J kg−1 and specific heat capacity of lead = 125 J kg1 K1.

500 m/s

5000 m/s

100 m/s

1000m/s

**Q.**

A lead bullet penetrates into a solid object and melts. Assuming that 50% of its kinetic energy was used to heat it, calculate the initial speed of the bullet. The initial temperature of the bullet is 27∘C and its melting point is 327∘C. Latent heat of fusion of lead = 2.5 × 104 J kg−1 and specific heat capacity of lead = 125 J kg1 K1.

500 m/s

5000 m/s

100 m/s

1000m/s

**Q.**Ice at 0∘C is added to 1 g of steam at 100∘C. For x g of ice added, the temperature of steam reduces to 0∘C. Find the value of x.

(Latent heat of ice =80 cal/g and latent heat of steam =540 cal/g. Also, specific heat of water =1 cal/ kg K)

**Q.**If two balls of the same metal weighing 5 gm and 10 gm strike a target with the same velocity and sticks to the target after the collision. The heat energy developed in the balls after striking is used for raising their temperatures, then the temperature will be

- higher for bigger ball
- higher for smaller ball
- equal for both the balls
- None of the above

**Q.**A 5 g piece of ice at - 20∘C is put into 10g of water at 30∘C. Assuming that heat is exchanged only between the ice and the water, find the final temperature of the mixture.

- 0∘C
- 10∘C
- 1000∘C
- 100∘C

**Q.**Figure shows three containers. The rightmost container contains water (heat capacity 5 cal/∘C), initially at temperature 100∘C. The middle container contains water maintained at 80∘C with the help of a heater. The container at the left hand side contains ice at 0∘C. There are two heat conducting rods A and B both having thermal resistance equal to 10∘C s/cal. Disregard any heat loss in the surrounding.

The temperature of water in the rightmost container decreases with time. The rate with which the ice melts in the leftmost container

- will decrease with time
- will increase with time
- will remain steady
- will increase and then decrease

**Q.**A bullet of mass 50 g and specific heat capacity 800 J kg−1 K−1 is initially at a temperature 20∘ C. It is fired vertically upwards with a speed of 200 m/s and on returning to the starting point strikes a lump of ice at 0∘ C and gets embedded in it. Assume that all the energy of the bullet is used up in melting. Neglect the friction of air. Latent heat of fusion of ice =3.36×105 J kg−1.

- Energy of bullet used in melting is 1800 J
- The mass of ice melted =5 g
- The mass of ice melted is slightly greater than 5 g
- The mass of ice melted is less than 5 g

**Q.**An ice cube of mass 0.1 kg at 0∘C is placed in an isolated container which is at 227∘C. The specific heat S of the container varies with temperature T according to the empirical relation S=A+BT, where A=100 cal/kg K and B=2×10−2 cal/kg K2. If the final temperature of the container (with the material in it) is 27∘C, determine the mass of the container.

[Latent heat of fusion of water =80 cal/g∘C, specific heat of water =1 cal/g∘C]

- 0.495 kg
- 0.495 g
- 4.950 kg
- 4.950 g

**Q.**How much heat must be removed by a refrigerator from 4 kg of water at 70∘C to convert it to ice cube at −10∘C ?

[Take specific heat of water Cw=4200 J/kg/∘C, Latent heat of fusion of iceLf=334000 J/kg, Specific heat of ice Ci=2100 J/kg K]

- 3521 kJ
- 2596 kJ
- 2352 kJ
- 4200 kJ

**Q.**How much heat energy is gained when 5 kg of water at 20∘C is brought to its boiling point?

(Specific heat of water=4.2 kJ kg−1c−1)

1680 kJ

1700 kJ

1720 kJ

1740 kJ

**Q.**Equal volumes of water and alcohol when put in similar calorimeters take 100 sec and 74 sec respectively, to cool from 50∘C to 40∘C. The thermal capacity of each calorimeter is numerically equal to the volume of either liquid. The specific gravity of alcohol is 0.8. If the specific heat capacity of water is 1 cal/g∘C, the specific heat capacity of alcohol will be -

[Assume constant rate of cooling]

- 0.6 cal/g∘C
- 0.8 cal/g∘C
- 1.6 cal/g∘C
- 1.8 cal/g∘C

**Q.**Figure shows three containers. The rightmost container contains water (heat capacity 5 cal/∘C), initially at temperature 100∘C. The middle container contains water maintained at 80∘C with the help of a heater. The container at the left hand side contains ice at 0∘C. There are two heat conducting rods A and B both having thermal resistance equal to 10∘C s/cal. Disregard any heat loss in the surrounding.

Initially, what is the power of heater required to maintain the temperature of middle container at 80∘C?

- 12 cal/s
- 6 cal/s
- 18 cal/s
- 20 cal/s

**Q.**22 g of CO2 at 27∘C is mixed with 16g of O2 at 37∘C. The temperature of the mixture is

- 27∘C
- 30.5∘C
- 32∘C
- 37∘C

**Q.**Work done in converting one gram of ice at –10∘C into steam at 100∘C is

- 3045 J
- 6056 J
- 721 J
- 616 J

**Q.**20 gm ice at −10 ∘C is mixed with m gm steam at 100 ∘C. The minimum value of m so that finally all ice and steam converts into water at 0 ∘C is

(Use: specific heat and latent heat as Cice=0.5 cal/gm ∘C, Cwater=1 cal/gm ∘C, Lfusion =80 cal/gm and Lvapor =540 cal/gm )

- 18527 gm
- 13517 gm
- 8532 gm
- 11317 gm

**Q.**

A cube of ice is floating in water contained in a vessel. When the ice melts, the level of water in the vessel

rises

falls

remains unchanged

falls at first and then rises to the same height as before.

**Q.**A block of ice at −10oC is slowly heated and converted to steam at 100oC. Which of the following curves represents the phenomenon qualitatively?

**Q.**The water equivalent of a copper calorimeter is 4.5 g. If the specific heat of copper is 0.09 cal g−1C−1, then

- Mass of the calorimeter is 0.5 kg.
- Thermal capacity of the calorimeter is 4.5 cal∘C−1.
- Heat required to raise the temperature of the calorimeter by 8∘C will be 36 cal
- Heat required to melt 15 gm of ice placed in the calorimeter will be 1200 cal