You have 60 kg of water at 273.2500K. The specific heat of water is 4186 J/K kg.

Question

You have 60 kg of water at 273.2500K. The specific heat of water is 4186 J/K kg. How much heat (in Joules) must be removed from the water to cool it all down to 273.1500K? Water has a latent heat of fusion of 333 kJ/kg. You have 90 g of ice at 273.1500K. How much heat (in Joules) does it take to melt all that ice? Show me how you find the final temperature (in mK)when the water and the ice combine. You must show your work. This problem is your opportunity to show me the physics concepts you know, take advantage of it, and include units.

Explanation / Answer

25

a)

m = 100 Kg

T = 273/25 K

specific heat , S = 4186 J

heat removed = m * S * (change in temperature)

heat removed = 100 * 4186 * (273.25 - 273.15)

heat removed = 41860 J

the heat removed is 41860 J

b)

let the heat needed is Q

Q = latent heat * mass

Q = 90 * 333 J

Q = 29970 J

the heat needed to melt the ice is 29970 J

c)

let the final temperture is Tf

(100 + 0.090) * 4186 * (Tf - 273.15 ) = 41860 - 29970

solving for Tf

Tf = 273.178 K

Tf = 273178 mK

the temperature of the mixture is 273178 mK

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