Question

please solve these please i need urgently please

Answer 1

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