please solve these please i need urgently please
Attachments:
horsey:
hey bro is it for the olympiad!
Answers
Answered by
0
11) energy required to convert ice into water (H1) = mLf
energy required to increase water temperature 0°C to 100° C ( H2) = mS∆T
energy required to change phase water to vapour ( H3) = mLv
now, total energy required = H1 +H2 + H3 =m ( Lf + S∆T + Lf)
=5( 80 +100+540)
=5(720)
=3600 cal
12) energy required = ms∆T + mLf + mS∆T = m(s∆T +Lf +S∆T)
=10(0.5×5 +80+ 1× 75)
=10(2.5 +80+75)
=25 + 800 +750
=1575 Cal
13) M kg of ice mixed with water of mass M kg 10°C
if water loose 10°c to 0°c then
H = 10M×10³ Cal
but ice want to convert in water
H" = M80× 10³ Cal
here H" > H
so, final temperature = 0° C
14) water loss maximum energy = 1× 1× 80 ( H) = 80 cal
ice have energy required to convert in water ( H")= 1×80 = 80 Cal
here H" = H
so, final temperature is 0°C
15) in the same way you solve it .
try it ownself then you understand concept
energy required to increase water temperature 0°C to 100° C ( H2) = mS∆T
energy required to change phase water to vapour ( H3) = mLv
now, total energy required = H1 +H2 + H3 =m ( Lf + S∆T + Lf)
=5( 80 +100+540)
=5(720)
=3600 cal
12) energy required = ms∆T + mLf + mS∆T = m(s∆T +Lf +S∆T)
=10(0.5×5 +80+ 1× 75)
=10(2.5 +80+75)
=25 + 800 +750
=1575 Cal
13) M kg of ice mixed with water of mass M kg 10°C
if water loose 10°c to 0°c then
H = 10M×10³ Cal
but ice want to convert in water
H" = M80× 10³ Cal
here H" > H
so, final temperature = 0° C
14) water loss maximum energy = 1× 1× 80 ( H) = 80 cal
ice have energy required to convert in water ( H")= 1×80 = 80 Cal
here H" = H
so, final temperature is 0°C
15) in the same way you solve it .
try it ownself then you understand concept
Similar questions
English,
7 months ago
Hindi,
7 months ago
Science,
1 year ago
Social Sciences,
1 year ago
Geography,
1 year ago