It is desired to transform 0.3 kg of ice at -20 oC into steam at 115 oC. The spe

Question

It is desired to transform 0.3 kg of ice at -20 oC into steam at 115 oC. The specific heat of ice is 2090 J/kg oC, the specific heat of liquid water is 4186 J/kg oC, and the specific heat of steam is 2010 J/kg oC. For water, the latent heat of fusion is 3.33 x 105 J/kg, and the latent heat of vaporization is 2.26 x 106 J/kg. Determine the heat required to: (a) raise the temperature of the ice from -20 °C to 0 °C. (4 points) (b) convert the ice into liquid water at 0 °C. (4 points) (c) raise the temperature of the liquid water from 0 °C to 100 °C. (4 points) (d) convert the liquid water into steam at 100 °C. (4 points) (e) raise the temperature of the steam from 100 oC to 115 °C. (4 points) If 2.00 x 105 J of heat is added to 0.3 kg of ice at -20 °C, what will be the final state (phase and temperature) of the system? (5 points) (f)

Explanation / Answer

a) Q1=mc× change in temperature = 0.3 × 2090 × 20= 12540J=0.1254×10^5

b) Q2= mL = 0.3×3.3×10^5= 0.99 ×10^5J

c) Q3 =mc (100-0)= 0.3 × 4186×100= 125580J

= 1.25580×10^5J

d)Q4= mL =0.3×2.26×10^6= 0.678×10^6J= 6.78×10^5J

e) Q5= mc×(115-100)=0.3× 2010×15= 9045J

f ) Q1+Q2= 1.1145×10^5J

(2-1.1145)×10^5= 0.8855×10^5

mc×(T-0) = 0.8855×10^5=88550

T= 88550/0.3×4186

T=70.51

Final phase will be water at temperature 70.51C

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