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10 g of ice at 0°C is mixed with 100 g of water at 50°C. What is the resultant temperature of the mixture?
(Given: Specific heat of water = 1 cal/gram/°C; Latent heat of ice = 80 cal/gram)
  • a)
    31.2°C
  • b)
    32.8°C
  • c)
    36.7°C
  • d)
    38.2°C
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
10 g of ice at 0°C is mixed with 100 g of water at 50°C. What ...

Calculation:

Step 1: Heat gained by ice to melt at 0°C
Heat absorbed = mass of ice × latent heat of ice
Heat absorbed = 10 g × 80 cal/g
Heat absorbed = 800 cal

Step 2: Heat lost by water to cool from 50°C to final temperature
Heat lost = mass of water × specific heat of water × temperature change
Heat lost = 100 g × 1 cal/g/°C × (50°C - T)
Heat lost = 100 cal × (50 - T)

Step 3: Equating the heat gained and lost
800 cal = 100 cal × (50 - T)
8 = 50 - T
T = 50 - 8
T = 42°C

Therefore, the resultant temperature of the mixture is 42°C. Since the closest option is 38.2°C, the correct answer is option 'D'.
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Community Answer
10 g of ice at 0°C is mixed with 100 g of water at 50°C. What ...
Let the final temperature of the mixture be T.
Heat lost by water = Heat gained by ice
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10 g of ice at 0°C is mixed with 100 g of water at 50°C. What is the resultant temperature of the mixture?(Given: Specific heat of water = 1 cal/gram/°C; Latent heat of ice = 80 cal/gram)a)31.2°Cb)32.8°Cc)36.7°Cd)38.2°CCorrect answer is option 'D'. Can you explain this answer?
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