Calorimetry
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(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
- 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
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
- 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
- 40 g
- 4 g
- 8 g
- 160 g
[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
Give an example, where high specific heat capacity of water is used as a heat reservoir.
- 9.1∘C
- 17.9∘C
- 27.2∘C
- 37.4∘C
- 0∘ C
- 10∘ C
- 20∘ C
- 30∘ C
[Take g=9.8 m/s2 and J=4.2 J/cal]
- 350 J
- 350 cal
- 375 J
- 375 cal
- 0∘C
- 40∘C
- 80∘C
- <0∘C
- The heat input
- The mass of the body
- The initial temperature
- The material it is made up of
- Volume of the system under consideration
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
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
(Latent heat of ice =80 cal/g and latent heat of steam =540 cal/g. Also, specific heat of water =1 cal/ kg K)
- higher for bigger ball
- higher for smaller ball
- equal for both the balls
- None of the above
- 0∘C
- 10∘C
- 1000∘C
- 100∘C
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
- 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
[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
[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
(Specific heat of water=4.2 kJ kg−1c−1)
1680 kJ
1700 kJ
1720 kJ
1740 kJ
[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
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
- 27∘C
- 30.5∘C
- 32∘C
- 37∘C
- 3045 J
- 6056 J
- 721 J
- 616 J
(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
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.
- 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