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Question Number 42262 by Necxx last updated on 21/Aug/18

Heat is supplied at a rate of 500W  to a pressure cooker containing  water and fitted with a safety   valve.Steam escape such that   water is lost at 12g/min.If the  heat is supplied at 900W,water is  lost at 20g/min.Calculate the  specific latent heat of steam.

$${Heat}\:{is}\:{supplied}\:{at}\:{a}\:{rate}\:{of}\:\mathrm{500}{W} \\ $$$${to}\:{a}\:{pressure}\:{cooker}\:{containing} \\ $$$${water}\:{and}\:{fitted}\:{with}\:{a}\:{safety}\: \\ $$$${valve}.{Steam}\:{escape}\:{such}\:{that}\: \\ $$$${water}\:{is}\:{lost}\:{at}\:\mathrm{12}{g}/{min}.{If}\:{the} \\ $$$${heat}\:{is}\:{supplied}\:{at}\:\mathrm{900}{W},{water}\:{is} \\ $$$${lost}\:{at}\:\mathrm{20}{g}/{min}.{Calculate}\:{the} \\ $$$${specific}\:{latent}\:{heat}\:{of}\:{steam}. \\ $$$$ \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 21/Aug/18

dQ=mcdT+(dm)L  (dQ/dt)=mc(dT/dt)+L(dm/dt)  (dQ_1 /dt)=mc(dT/dt)+L(dm_1 /dt)  (dQ_2 /dt)=mc(dT/dt)+L(dm_2 /dt)  L=(((dQ_2 /dt)−(dQ_1 /dt))/((dm_2 /dt)−(dm_1 /dt)))=((900−500)/(((20×10^(−3) )/(60))−((12×10^(−3) )/(60))))  L=((400×60)/(10^(−3) ))×(1/8)=3×10^6 Joule  pls check...

$${dQ}={mcdT}+\left({dm}\right){L} \\ $$$$\frac{{dQ}}{{dt}}={mc}\frac{{dT}}{{dt}}+{L}\frac{{dm}}{{dt}} \\ $$$$\frac{{dQ}_{\mathrm{1}} }{{dt}}={mc}\frac{{dT}}{{dt}}+{L}\frac{{dm}_{\mathrm{1}} }{{dt}} \\ $$$$\frac{{dQ}_{\mathrm{2}} }{{dt}}={mc}\frac{{dT}}{{dt}}+{L}\frac{{dm}_{\mathrm{2}} }{{dt}} \\ $$$${L}=\frac{\frac{{dQ}_{\mathrm{2}} }{{dt}}−\frac{{dQ}_{\mathrm{1}} }{{dt}}}{\frac{{dm}_{\mathrm{2}} }{{dt}}−\frac{{dm}_{\mathrm{1}} }{{dt}}}=\frac{\mathrm{900}−\mathrm{500}}{\frac{\mathrm{20}×\mathrm{10}^{−\mathrm{3}} }{\mathrm{60}}−\frac{\mathrm{12}×\mathrm{10}^{−\mathrm{3}} }{\mathrm{60}}} \\ $$$${L}=\frac{\mathrm{400}×\mathrm{60}}{\mathrm{10}^{−\mathrm{3}} }×\frac{\mathrm{1}}{\mathrm{8}}=\mathrm{3}×\mathrm{10}^{\mathrm{6}} {Joule}\:\:{pls}\:{check}... \\ $$

Commented by Necxx last updated on 21/Aug/18

Yeah....Its correct sir..Thanks

$${Yeah}....{Its}\:{correct}\:{sir}..{Thanks} \\ $$

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