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19g of water at 30°C and 5g of ice at − 20°C are mixed together in a calorimeter. What is the final temperature of the mixture ? Given specific heat of ice = 0.5 cal g − 1(°C)− 1 and latent heat of fusion of ice = 0.8 cal g − 1
  • a)
    0°C
  • b)
    − 5°C
  • c)
    5°C
  • d)
    10°C
Correct answer is option 'C'. Can you explain this answer?
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19g of water at 30°C and 5g of ice at − 20°C are mixed t...
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19g of water at 30°C and 5g of ice at − 20°C are mixed t...
°C is heated until its temperature reaches 100°C. The specific heat capacity of water is 4.18 J/g°C.

To find the amount of heat gained by the water, we can use the formula:

Q = m * c * ΔT

Where:
Q = heat gained by the water (in Joules)
m = mass of the water (in grams)
c = specific heat capacity of water (in J/g°C)
ΔT = change in temperature (in °C)

Given:
m = 19g
c = 4.18 J/g°C
ΔT = (100°C - 30°C) = 70°C

Substituting these values into the formula:

Q = 19g * 4.18 J/g°C * 70°C
Q = 5266.6 J

Therefore, the amount of heat gained by the water is 5266.6 Joules.
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19g of water at 30°C and 5g of ice at − 20°C are mixed together in a calorimeter. What is the final temperature of the mixture ? Given specific heat of ice = 0.5 cal g − 1(°C)− 1 and latent heat of fusion of ice = 0.8 cal g − 1a)0°Cb)− 5°Cc)5°Cd)10°CCorrect answer is option 'C'. Can you explain this answer?
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