How much energy is absorbed when 10g of ice at 0 C becomes steam at 100C
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Latent heat of fusion of ice (L) = 80 cal/g
Latent heat of vapourisation of water (L’) = 540 cal/g
Specific heat of water (S) = 1 cal/(g °C)
Energy absorbed = mL + mSdT + mL’
= m (L + SdT + L’)
= 10 g × (80 + (1 × 100) + 540)
= 7200 calories
= 7.2 kcal
Heat absorbed is 7.2 kcal
Latent heat of vapourisation of water (L’) = 540 cal/g
Specific heat of water (S) = 1 cal/(g °C)
Energy absorbed = mL + mSdT + mL’
= m (L + SdT + L’)
= 10 g × (80 + (1 × 100) + 540)
= 7200 calories
= 7.2 kcal
Heat absorbed is 7.2 kcal
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