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The water equivalent of a calorimeter is 10 g and it contains 50 g of water at 15^@ C. Some amount of ice, initially at -10^@ C is dropped in it and half of the ice melts till equilibrium is reached. What was the initial amount of ice that was dropped (when specific heat of ice = 0.5 cal g^(-1) ""^(@) C^(-1), specific heat of water = 1.0 cal g^(-1) ""^(@) C^(-1) and latent heat of melting of ice = 80 cal g'^(-1))? |
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Answer» 10 g  Massofwaterequivalent ofa calorimeterand watercontinedby it `m_1 = 10G+ 50 g =60 g ` the amount of heatlost bythe water ` Q_1 = m_1s Delta T ` wheres=specificheatof water,` DeltaT =`temperature ` thereforeQ_1( 60 xx 1xx 15)`cal AMOUNTOF heatgainedby iceis `Q_2 =(m )/(2) xx10 + (m)/(2)xx 80 = (45m)`cal in steadystateheatgained`= `heatlost `therefore45m= 60xx 1 xx 15 implies m -20g` |
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