You have 110 kg of water at 273.2500K. The specific heat of water is 4186 J/K kg

Question

You have 110 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 50 g of ice at 273.1500K. How much heat (in Joules) does it take to melt all that ice? Show me how you find how much ice (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

a) Specific heat capacity of water is 4181J / kg degrees C

So to take 110kg of water and remove 273.1500K = -0.05 degress C from it will take:

110kg * (-0.05) * 4181 J/kg°C = 22995.5 Joules or 22.9955 kJ

b) since the ice cube is at -0.05 degrees celcius

Q = Hf * m
Hf = 3.33 * 10^5 J/kg
Q = 3.33 * 10^5 * 0.05 kg = 1.6650 * 10^3 J

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