10 gram of ice at 0 degree c absorbs 5460 j heat energy to melt and change to water at 50 degree
c. calculate the specific latent
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Heat required for melting=mLf
where m is mass of ice and Lf is Latent heat of fusion.
and heat required to change to the temperature of water from 0°C to 50°C =mS (50-0)=50mS
where S is specific heat of water.
We know,
Heat Loss=Heat gain,
5460 Joule=mLf+50mS
5460=10×10^-3 Lf+50×10×10^-3×420p
5460-2100=10^-2Lf
Lf=3360×10 to the power 2 =3.36×10^5j/kg
where m is mass of ice and Lf is Latent heat of fusion.
and heat required to change to the temperature of water from 0°C to 50°C =mS (50-0)=50mS
where S is specific heat of water.
We know,
Heat Loss=Heat gain,
5460 Joule=mLf+50mS
5460=10×10^-3 Lf+50×10×10^-3×420p
5460-2100=10^-2Lf
Lf=3360×10 to the power 2 =3.36×10^5j/kg
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