When 0.15kg of ice at 0oC is mixed with 0.30kg of water at 50o C in a container, the resulting temperature is 6.7o C. Calculate latent heat of melting of ice.(Specific heat of water is 4.186×103 JKg−1).
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Answers
Heat lost by water = mSw (θf - θi)w = (0.30 kg) (4186 J kg-1°C-1) x (50.0°C - 6.7°C) = 54376.14 J Heat to melt ice = m2Lf = (0.15 kg)Lf Heat to raise temperature of ice water to final temperature = m1Sw (θf - θi) = (0.15 kg) (4188 J kg-1°C-1) (6.7°C - 0°C) = 4206.93 J Heat lost = Heat gained 54.76.14 J = (0.15 kg) Lf + 4206.93 J ∴ Lf = 3.34 x 105 Jkg-1.Read more on Sarthaks.com - https://www.sarthaks.com/562994/when-of-ice-of-0c-mixed-with-30-kg-water-at-50c-in-container-the-resulting-temperature-7c
Answer:
⇒ΔTwater=(273.15+50)−(273.15+6.7)
⇒ΔTwater=43.3K
Substitute 0.30kgfor the mass of water, 4186J.kg−1K−1
for the heat capacity of water and 43.3K for temperature change.
qwater=0.30kg×4186J.kg−1K−1×43.3K
qwater=54376.14J
Explanation:
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