A burner supplies heat energy at the rate of 434 J/s for 60 s when 40 g of ice at 0°c changes to water at 75°c.Calculate the latent heat of ice?
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Answered by
10
Q = Heat supplied = 434 J/s * 60s = 26,040 J
Latent heat of fusion of ice = L J / gm
Specific heat of water = 4.18 J /°C/gm
Heat absorbed by ice to become water at 0°C = H1 = 40 * L joules
Heat absorbed by water at 0 °C to 75°C = 4.18 * 75° * 40 gm
H2 = 12,540 J
Q = H1 + H2
26, 540 = 40 L + 12,540
L = 350 J/gm
Latent heat of fusion of ice = L J / gm
Specific heat of water = 4.18 J /°C/gm
Heat absorbed by ice to become water at 0°C = H1 = 40 * L joules
Heat absorbed by water at 0 °C to 75°C = 4.18 * 75° * 40 gm
H2 = 12,540 J
Q = H1 + H2
26, 540 = 40 L + 12,540
L = 350 J/gm
Answered by
6
heat loss = heat gain
heat loss = 434 x 60 Joule
heat gain = mLf + ms∆T
now,
434 x 60 = m( Lf + 4200 x 75)
434 x 60 = 40 x 10^-3 ( Lf + 75 x 4200)
434 x 60 x 10³ /40 = Lf + 75 x 4200
Lf = 336000 j/kg
=3.36 x 10^5 j/kg
heat loss = 434 x 60 Joule
heat gain = mLf + ms∆T
now,
434 x 60 = m( Lf + 4200 x 75)
434 x 60 = 40 x 10^-3 ( Lf + 75 x 4200)
434 x 60 x 10³ /40 = Lf + 75 x 4200
Lf = 336000 j/kg
=3.36 x 10^5 j/kg
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