Last Updated On 01-06-2023 - By Jeesha Sara
Ice at −20°c is added to 50 g of water at 40° c. when the temperature of the mixture reaches 0°c, it is found that 20 g of ice is still un melted. The amount of ice added to the water was close to (specific heat of water =4.2 J/g/°c) Specific heat of Ice=2.1 J/g/°C Heat of fusion of water at 0°C = 334 J/g)
(a)
50g
(b)
40g
(c)
60g
(d)
100g
Ice at −20°c is added to 50 g of water at 40° c. when the temperature of the mixture reaches 0°c, it is found that 20 g of ice is still un melted. The amount of ice added to the water was close to (specific heat of water =4.2 J/g/°c) Specific heat of Ice=2.1 J/g/°C Heat of fusion of water at 0°C = 334 J/g)
(a)
50g
(b)
40g
(c)
60g
(d)
100g
Question ID - 50308 :-
Ice at −20°c is added to 50 g of water at 40° c. when the temperature of the mixture reaches 0°c, it is found that 20 g of ice is still un melted. The amount of ice added to the water was close to (specific heat of water =4.2 J/g/°c) Specific heat of Ice=2.1 J/g/°C Heat of fusion of water at 0°C = 334 J/g)
(a)
50g
(b)
40g
(c)
60g
(d)
100g
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Ice at −20°c is added to 50 g of water at 40° c. when the temperature of the mixture reaches 0°c, it is found that 20 g of ice is still un melted. The amount of ice added to the water was close to (specific heat of water =4.2 J/g/°c) Specific heat of Ice=2.1 J/g/°C Heat of fusion of water at 0°C = 334 J/g) |
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|
(a) |
50g |
(b) |
40g |
|
(c) |
60g |
(d) |
100g |
Let amount of ice is m gm.
According to principal of calorimeter
Heat taken by ice=heat given by water
∴20×2.1×m+(m−20)×334 = 50×4.2×40
376 m=8400+6680
M=40.1
∴correct answer is (b)
Next Question :
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In the figure shown below, the charge on the left plate of the 10
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(a) |
−18 μC |
(b) |
−12 μC |
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(c) |
+12μC |
(d) |
+18 μF |
capacitor is −30
? The Charge on the right plate of the 6 
capacitor is
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