calculate the totl amount of heat energy required to convert 100gm of ice at- 10 °C completely into watr at 100°C . specfc heat capacity of ice is 2.1 J per gm per Kelvin... specfc heat capacity of watr is 4.2 j per gm per Kelvin ... specfc latent heat of ice is 336 j per gm ... plzzz slv it fast friends......
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77700 Joules
77.7 KiloJoules
77.7 KiloJoules
hrik2:
why hv u only writen the ansr .. ansr i know.. u shud write the ful solution
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heat gain by ice to rise its temp upto 0c=mass of ice ×specific heat capacity of ice× temp change
= 100×2.1×10=2100 J
heat gain to turn into water=its mass ×its latent heat of fusion
=100×336=33600J
heat gain to rise temp upto 100c=its mass× specific heat capacity of water× temp change
=100× 4.2× 100=42000 J
total heat gain=
2100+33600+42000
=77700J= 77.7 kJ
= 100×2.1×10=2100 J
heat gain to turn into water=its mass ×its latent heat of fusion
=100×336=33600J
heat gain to rise temp upto 100c=its mass× specific heat capacity of water× temp change
=100× 4.2× 100=42000 J
total heat gain=
2100+33600+42000
=77700J= 77.7 kJ
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