1) heat lost by metal = heat gained by water
       Â
m1*s1*DT1 = m2*s2*DT2
m1 = mass of metal = 50 g
s1 = specific heat of metal = ?
DT = 100-24 = 76 c
m2 = mass of water = 25 g
s2 = specific heat of water = 4.184 j/g.c
DT2 = 24-10 = 14
50*S1*76 = 25*4.184*14
S1 = specific heat of metal = 0.385 j/g.c
answer: c.0.385 j/g.c
2)
according to reaction: 1 mol CH4 = 890.4 Kj
SO That, No of mol of CH4 required =
1000/890.4
                                   Â
= 1.123 mol
amount of CH4 required = n*Mwt
                     Â
= 1.123*16
                     Â
= 17.968 g
                     Â
= 18 g
answer: d. 18 g
