If one gram of kerosene liberates 46 kJ of heat when it is burned to what temperature can 0.25 g of kerosene raise the temperature of 75 cm 3 of water at 25° C ?

Question

If one gram of kerosene liberates 46 kJ of heat when it is burned to what temperature can 0.25 g of kerosene raise the temperature of 75 cm 3 of water at 25° C ? (A) 61.6° C (B) 46° C (C) 66.1° C (D) 64° C

Tardigrade Admin · Accepted Answer

Heat liberated =(1/4) × 46=11.5 =((11500/4.184)) cal Hence (75 g )(1 cal ° C -1 g -1)( t -25° C )= cal t -25° C =36.65° C

Q. If one gram of kerosene liberates $46 k J$ of heat when it is burned to what temperature can $0.25 g$ of kerosene raise the temperature of $75 c m^{3}$ of water at $2 5^{\circ} C$ ?

$61. 6^{\circ} C$

$4 6^{\circ} C$

$66. 1^{\circ} C$

$6 4^{\circ} C$

Heat liberated $= \frac{1}{4} \times 46 = 11.5$
$= (\frac{11500}{4.184}) c a l$
Hence $(75 g) (1 c a l^{\circ} C^{- 1} g^{- 1}) (t - 2 5^{\circ} C) = c a l$
$t - 2 5^{\circ} C = 36.6 5^{\circ} C$

Q. If one gram of kerosene liberates 46kJ of heat when it is burned to what temperature can 0.25g of kerosene raise the temperature of 75cm3 of water at 25∘C ?

61.6∘C

46∘C

66.1∘C

64∘C