how many joules of heat is absorbed by water at 0°C
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the heat(Q- in joules) needed to raise the temperature of the sample of water of mass M(in grams) by T(º celsius) is.....
Cp= specific heat of the water (4.18 J/g*ºC)
Q=M*Cp*∆T
So....
V=1000mL so M=1000mL*(1g/1mL)= 1000g water
Q=(1000g)(4.18J/g*ºC)(50-25)=104500 J or 104.5 kJ
Cp= specific heat of the water (4.18 J/g*ºC)
Q=M*Cp*∆T
So....
V=1000mL so M=1000mL*(1g/1mL)= 1000g water
Q=(1000g)(4.18J/g*ºC)(50-25)=104500 J or 104.5 kJ
Nikhil131Chopra32:
sry u didn't understand the question
You may please refer to my answer
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When ice melts at 0 degrees C it draws 3.34×10^5 joule/kg heat from the substance it touches to get converted into water ... then the water at 0 degrees C draws or absorbs in 22.5 × 10^5 joule/kg heat from the substance it touches in order to vaporize and gets converted into water vapour.
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