18. How much heat must be removed from 0.15 kg of water to cool it from 95°C to

Question

18. How much heat must be removed from 0.15 kg of water to cool it from 95°C to 25 "C? The specific heat capacity of water is 4182 J/kgK. 19. A piece of copper (specific heat 385 J/kg K) has a mass of 275 g. If it is heated to 145°C, then plunged into 5.00 kg water (specific heat-4 180 J/kg-K) at 200, what will be the final temperature at equilibrium? 500 373 150 0 Heat (J) He=3.34 x 105 J/kg H-2.26 x 108 Jkg C(ice)- 2060 x J/kg K C(water)- 4180 x J/kg K C(steam) 2020 x J/kg K 20. Find the amount of energy required to change a 4,50 kg block of ice just at its melting point to a gas at the temperature shown by point g on the graph.

Explanation / Answer

Calculate the heat removed to cool water.

E = mw * cw * t

E = 0.15 kg * 4182J/kg k * (95-25)

E = 0.15 kg * 4182J/kg k * (70k)

= 43911 J = 43.911kJ

so, in order to cool the water from 95 to 25C, 43.9kJ of heat must be removed.

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