When same quantity of heat is supplied to 300 gm of water, 150 gm of oil. rise in water 10 C, OIL 40 C. Calculate Q and S.
1 Answer
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
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