Calorimetry
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1 kg of ice at 0∘C is mixed with 1 kg of steam at 100∘C. What will be the composition of the system when thermal equilibrium is reached? Latent heat of fusion of ice = 3.36 × 105 J kg−1, specific heat capacity of water = 4200 J kg −1 K−1 and latent heat of vaporization of water = 2.26 × 106 J kg−1
2kg steam
2 kg ice
665gm water, 1.335 kg steam
665 gm steam, 1.335 kg water
(Specific heat of water =1kcalkg−1, specific heat of ice =0.50kcalkg−1 while its latent heat =80kcalkg−1)
- Ice is 16.75 gm , and water is 11.25 gm at .
- Ice is 6.75 gm , and water is 181.25 gm at .
- Ice is 108.75 gm , and water is 101.25 gm at .
- Ice is 68.75 gm , and water is 181.25 gm at .
- 20∘ C
- 30∘ C
- 0∘ C
- 10∘ C
- 0.625 g ice will melt. Temperature will remain 0oC
- 0.625 g ice will melt. Temperature will increase to 5oC
- 10 g ice will melt. Temperature will increase to 5oC
- 10 g ice will melt. Temperature will remain 0oC
passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to
the boiler as water at 90∘C. how many kg of steam is required per hour.
(Specific heat of steam = 1 calorie per gm°C, Latent heat of vaporization = 540 cal/gm)
- 1 gm
- 1 kg
- 10 gm
- 10 kg
- 1.5×107J
- 3.5×107J
- 2.5×107J
- 4.5×107J
- 6.25 g ice will melt. Temperature will remain 0∘
- 0.625 g ice will melt. Temperature will increase to 10∘
- 0.625 g ice will melt. Temperature will remain 0∘
- 6.25 g ice will melt. Temperature will increase to 10∘
Steam at 100∘C is passed into 1.1 kg of water contained in a calorimeter of water equivalent 0.02 kg at 15∘C till the temperature of the calorimeter and its contents rises to 80∘C. The mass of the steam condensed in kg is
0.130
0.065
0.260
0.135
- 273 K
- 373 K
- 300 K
- 0 K
- 3.03×106 J
- 4.03×106 J
- 5.03×106 J
- 5.03×106 J
On which of the following does the specific heat capacity of a body depend?
The heat impart
The mass of the body
Temperature
The material it is made up of
- 2.55J
- 25.5J
- 255J
- 25, 500J