Calorimetry
Trending Questions
- 116.4∘C
- 216.4∘C
- 110.4∘C
- 126.4∘C
- BC represents the change of state from solid to liquid
- T2 is the melting point of the solid
- (H2−H1) represents the latent heat of fusion of the substance
- (H3−H1) represents the latent heat of vaporization of the liquid
Heat required to convert one gram of ice at 0∘C into steam at 100∘C is (given Lstream = 536 cal/gm)
1 kilo calorie
100 calorie
0.01 kilo calorie
716 calorie
- 0∘ C
- 10∘ C
- 20∘ C
- 30∘ C
- 24.7×104 J/kg
- 210.4×104 J/kg
- 2.4×104 J/kg
- 12.4×104 J/kg
(Specific heat of water=4.2 kJ kg−1c−1)
1680 kJ
1700 kJ
1720 kJ
1740 kJ
- 0.1
- 1
- 10
- None of these
- 27∘C
- 30.5∘C
- 37∘C
- 32∘C
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
- 3045 J
- 6056 J
- 721 J
- 616 J
A student heats a beaker containing ice and water. He measures the temperature of the content of the beaker as a function of time. Which of the following would correctly represent the result ? Justify your choice.
A
B
C
D
and the resulting temperature is found to be 35∘C. The thermal capacity of the calorimeter is
- 6300 J/K
- 1260 J/K
- 4200 J/K
- None of these
(Cice=0.5 k cal kg−1k−1, Lice=80 k cal kg−1)
- 31.25 g
- 150 g
- 210 g
- 181.25 g
- Q4>Q3>Q2>Q1
- Q4>Q3>Q1>Q2
- Q4>Q2>Q3>Q1
- Q4>Q2>Q1>Q3
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
- 600J/∘C, 15J/kg∘C
- 150J/∘C, 60J/kg∘C
- None of these
- 15J/∘C, 600J/kg∘C
- Mass of ice melted = Mass of water vaporised
- Mass of ice melted < Mass of water vaporised
- Can't be determined.
- Mass of ice melted > Mass of water vaporised
A mass of 50 g of a certain metal at 150∘C is immersed in 100 g of water at 11∘C. The final temperature is 20 ∘C. Calculate the specific heat capacity of the metal. Assume that the specific heat capacity of water is 4.2 J g−1 K−1.
(J = 4.2 Joule/cal.)
- 22.14∘C
- 42.14∘C
- 142.14∘C
- 420.14∘C