Question

(a) Calculate the amount of heat given out while 400 g of water at 30 °C is cooled and converted into ice at -2 °C.

Specific heat capacity of water = 4200 J/kg K

Specific heat capacity of ice = 2100 J/ kg K

Specific latent heat of fusion of ice = 336000 J/kg

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Answer 1

Answer:

Heat given by the water is 186480 J

Explanation:

As we know when water is cooled down then heat is rejected by it

The total amount of heat rejected by water when its temperature is cooled down is given as

$Q = ms\Delta T$

similarly the amount of heat rejected by water when it is converted into ice is given as

$Q = mL$

so we will have

$Q = ms\Delta T + mL$

$Q = 0.400(4200)(30 - 0) + 0.400(336000) + 0.400(2100)(0 - (-2))$

$Q = 50400 + 134400 + 1680$

$Q = 186480 J$

#Learn

Topic : Calorimetry

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