gIVEN,
FOR WATER,
MASS (m1) = 300 GM = 0.3 KG
RISE IN TEMPERATUREdt1 = 10 °C
specific heat capacity of water s1 = 4200 J/kg°C
Now, heat supplied
Q = m1* s1* dt1
= 12600 J
fOR OIL
mas m2 = 150 gm = 0.15 kg
rise in temp (dt2) = 40 °C
heat supplied Q = 12600 J
specific heat of oil s2 = ??
again,
Q = m2 *s2 * dt2
S2 = 12600 / ( 0.15 * 40 )
S2 = 2100 j/ KG°C