A 5 kg piece of lead (specific heat 0.03 cal/g/ degree C) having a temperature o

Question

A 5 kg piece of lead (specific heat 0.03 cal/g/ degree C) having a temperature of 80 degree C is added to 500 g of water having a temperature of 20 degree C. What is the final equilibrium temperature (in degree C) of the system? We consider the physical of phase changes of a 1kg of ice at -20 degree C in a container that is held at constant pressure to steam. The figure below shows the temperature of the material as a function of the added heat. (Latent heat of fusion of water is 3.33 Times 10^5 J/kg); (Latent heat of vaporization of water is 2.26 Times 10^6 J/kg); (specific heat of ice is 2090 J/kg. degree C); (specific heat of water is 4186 J/kg. degree C) What is the heat energy needed to raise temperature of the ice from -20 to 0 degree C (kJ)? What is the heat energy needed to melt 1 kg of ice (kJ)? What is the heat energy needed to raise temperature of the resulting 1kg of water from 0 to 100 degree C(kJ)? What is the heat energy needed to vaporize 1 kg of water at 1

Explanation / Answer

Q21. specific heat of water=4.186 cal/g/degree celcius

as there are no heat loss or heat adition,

heat loss by lead=heat gain by water

let final temperature of the system be T degree celcius.

==>mass of lead*specific heat of lead*temperature difference=mass of water*specific heat of water*temperature difference

==>5000*0.03*(80-T)=500*4.186*(T-20)

==>10*0.03*(80-T)=4.186*(T-20)

==>24-0.3*T=4.186*T-83.72

==>24+83.72=(4.186+0.3)*T

==>T=(24+83.72)/(4.186+0.3)

==>T=24.012 degree celcius

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