how much heat is required to melt 60g of ice at its normal temperature
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Answer:
How much heat is needed to convert 80g of ice initially at -10 degree celsius to 60 g of water and 20 g of steam at 100 degree celsius?
Solving this problem would require both sensible and latent heat calculations. Overall heat balance can be written as
Q(total)=Q1+Q2+Q3+Q4
where Q(total) = Total heat required
Q1 heat required to increase the temperature of ice from -10 °C to 0°C
Q2 Heat required for conversion of ice at 0 °C to water at 0 °C (Latent heat of fusion)
Q3 Heat required to increase the temperature of water from 0°C to 100 °C
Q4 Heat required for vaporization of water at 100 °C (Latent heat of vaporization)
Q(total)=m1.Cp(ice)ΔT1+m1.λ(fusion)+m1.Cp(water).ΔT2+m2.λ(vaporization)
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How much heat is necessary to change 500 g of ice at -10 degrees Celsius to water at 20 degrees Celsius?
How much heat is needed to convert 10 grams of ice at 0°C to water at 25°C?
How much heat is required to bring 1.0 kg of water from 25 degrees Celsius to 99 degrees Celsius?
What is the amount of heat required to convert one gram of ice at zero degree Celsius to water at 20 degrees Celsius?
Is the heat required to convert 75.0 g. of ice at 0.0 degrees Celsius to steam at 100.0 degrees Celsius?
80 gms of ice at -10℃ converted to 80 gms of ice at 0℃ , heat consumed =80×0.5{0-(-10)} cal =80×0.5×10 = 400 cal ……..(1)
80 GMs of ice will melt at 0℃ , heat consumed
=80×80 =6400 cal………..(2)
80 gms of water will be heated from 0℃ to 100℃
Heat consumed =80×1×(100–0)
=8000 cal……….(3)
Heat consumed for vaporisation of 20 gms of water=20×540 =10800 cal ….(4)
Total heat consumed =(1)+(2)+(3)+(4)=400+6400+8000+10800 = 25600 cal
Answer:
answer of that is 12 only