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19 g of water at 30° C and 5 g 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 = 80 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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19 g of water at 30° C and 5 g of ice at – 20° C are mix...
°C is mixed with 31 g of water at 50°C. What is the final temperature of the mixture?

To solve this problem, we need to use the principle of conservation of energy:

Q1 + Q2 = Q3

where Q1 is the heat absorbed by the first sample of water, Q2 is the heat absorbed by the second sample of water, and Q3 is the heat released by the final mixture.

We can calculate Q1 and Q2 using the specific heat capacity of water:

Q1 = 19 g x 4.18 J/g·°C x (Tf - 30°C)
Q2 = 31 g x 4.18 J/g·°C x (Tf - 50°C)

where Tf is the final temperature of the mixture.

We can simplify these equations by factorizing out the specific heat capacity and solving for Tf:

Q1 = Q2
19 x (Tf - 30) = 31 x (50 - Tf)
19Tf - 570 = 1550 - 31Tf
50Tf = 2120
Tf = 42.4°C

Therefore, the final temperature of the mixture is 42.4°C.
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19 g of water at 30° C and 5 g of ice at – 20° C are mix...
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19 g of water at 30° C and 5 g 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 = 80 cal g–1a)0° Cb)– 5° Cc)5° Cd)10° CCorrect answer is option 'C'. Can you explain this answer?
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