How much energy, in k3, is needed to melt 56.8 grams of ice at 0.00 °c to water

Question

How much energy, in k3, is needed to melt 56.8 grams of ice at 0.00 °c to water at 0.00 °c Problem 3 How much energy, in k), is needed to heat 21.4 grams of ice at -97.3 °C to ice at -5.26C k) Tries 0/5 Problem 4 How much energy, in kJ, is needed to convert 65.5 grams of ice at o.00 °c to water at 70.1 °c? Tries o/5 Problems How much energy, in kJ, is needed to convert 22.9 grams of water at 9.17 c to vapor at 147 ka it Answer Tries 0/s Problem 6 How much energy, in kJ, is needed to convert 38.2 grams of water at -20.6c to vapor at 142 2 Tries 0/s s discussion is closed

Explanation / Answer

1) Q = m Lf

m =mass of water = 56.8g, Lf = latent haet of fusion = 334 J/g

Q = 56.8 g x 334 J / g = 18971 J = 18.97kJ

2) Q =Si m dT

Si = specific heat of Ice = 2.1 J / g.K ==> heat req. to raise 1 g ice temp by 1 K

dT = [ (273.15 + (-5.26oC)) - (273.15 +(-97.3oC)) ] = - 92.04 K ( minus is negated only quantity considered)

m = 21.4g

Q = 2.1 J / g.K x 21.4g x 92.04K = 4136 J = 4.14 kJ

3) Q = m Lf + Sw m dT ( Include melting of ice at 0 oC and heating from 0oC to 70.1 oC )

Sw = Specific heat of water = 4.18 J/ g.K ==> heat req raise 1 g of water by 1K

dT = (273.15K + 70.1oC) - (273.15K+0oC) = 70.1 K

m = 65.5g

Lf = 334 J/g

Q = m Lf + Sw m dT

= (65.5g x 334 J /g) + ( 4.18 J/ g.K x 65.5g x 70.1 K )

= 21877 J + 19192.6 J

= 41069.6 J

= 41kJ

Rest upload as new question or follow below instruction to solve the answer

4) Just add Q = Sw m dTw + mLv + Ss m dTv

m =22.9g, Ss(steam) = 1.996 J/g.K, Lv = 2.26 J /g, m = mass, dtw = 90.83K, dTv = 47K

5) Q = Si m dTi + m Lf + Sw m dT + m Lv + Ss m dTs

follow the instructions as above

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