Albert makes a mug of tea but it is too hot to drink. The mug and the tea have a

Question

Albert makes a mug of tea but it is too hot to drink. The mug and the tea have an initial temperature of 97 degree C and he wants to cool them down to 87 degree C using ice cubes. The tea (mostly water) has a mass of 0.25 kg and the ceramic mug has a mass of 0.36 kg. If each ice cube has a mass of 0.0036 kg and an initial temperature of 0 degree C, how many ice cubes will he need? (Assume he is using a well-insulated container) Constants: Specific heat of ceramic: 850 J/(kg*K) Specific heat of water: 4190 J/(kg*K) Specific heat of ice: 2220 J/(kg*K) Latent heal of fusion for water: L_F = 334000 J/kg Solution:

Explanation / Answer

Heat removed from ceramic and tea system Q = MC(97-87) +mc(97-87)

Q = [0.25x4190x10]+[0.36x850x10]

= 13535 J

Heat agin by each ice cube Q ' = m ' Lf + m' C (87-0)

= 0.0036(334000) +0.0036(4190)(87)

= 2514.708 J

Number of ice cubes N = Q / Q '

= 5.382

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