find the amount of heat required to change 10gram of ice at -5℃ to water at 20℃????
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First the ice at - 5°C will convert into 0°C
So heat required will be :
Q¹=m×S(ice) ×(temp. Change)
=10× 0.5×(0-(-5))
=25cal
[S(ice) =specific heat of ice=0.5cal/g-°C]
Now ice will change its state and melt to become water so heat required will be
Q²=m×L(ice)
=10×80=800cal
[ L(ice) =latent heat of fusion of ice=80cal/g]
Now heat required to raise the temp by 20°Cwill be
Q³=m×S(water) ×(temp change)
=10×1×(20-0)
=200cal
[S(water) = specific heat of water =1cal/g-°C]
Total Q=25+800+200=1028cal
So heat required will be :
Q¹=m×S(ice) ×(temp. Change)
=10× 0.5×(0-(-5))
=25cal
[S(ice) =specific heat of ice=0.5cal/g-°C]
Now ice will change its state and melt to become water so heat required will be
Q²=m×L(ice)
=10×80=800cal
[ L(ice) =latent heat of fusion of ice=80cal/g]
Now heat required to raise the temp by 20°Cwill be
Q³=m×S(water) ×(temp change)
=10×1×(20-0)
=200cal
[S(water) = specific heat of water =1cal/g-°C]
Total Q=25+800+200=1028cal
Vanshiiitaaa:
Welcome
when u convert 1025 calories into joules u will get 4300 j
From which book this question is?
If you take the values of specific heat and latent heat in joules you will get that answer because in cal calculations become simple so i took it in calories. For example : specific heat of water in cal is 1 but in joules it is 4.186
this makes the difference of 5
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