Question #130254
A carnot engine has its source at 120°C and its sink is maintained at a constant temperature by means of ice at 0°C. If it is working at the rate of 175 watts, how much ice will melt in two minutes?
1
Expert's answer
2020-08-21T10:59:20-0400

Given data

Temperature of source (T(Tsource)=1200C+273

=393 K

Temperature of sink (Tsink)=00C+273

=273 K

Working rate (power) =175 watts

Workdone by carnot engine in 2 minutes

W=P×tW=P\times t

W=175×2×60W=175\times2 \times60

W=21000JW=21000 J

efficiency of carnot engine

e=1TsinkTsourcee=1-\frac{T_{sink}}{T_{source}}

e=1273393e=1-\frac{273}{393}


e=120393=0.305e=\frac{120}{393}=0.305

efficiency of carnot engine can be given by formula mention below

e=WQe=\frac{W}{Q} Where Q is heat rejected at sink

therefore Q=WeQ= \frac{W}{e}


Q=210000.305Q=\frac{21000}{0.305}


Q=6405JQ=6405J

This rejected heat will melt the ice and can be given by below

Q=mLfQ={m}{L_{f}}

where m is the mass of melted ice and Lf is latent heat of fusion.

therefore

m=QLfm=\frac{Q}{L_f} value of latent heat of fusion for

ice Lf =3.33×105J/Kg3.33\times10^5 J/Kg


by putting the value of Q annd Lf


m=64053.33×105m=\frac{6405}{3.33\times 10^5}

m=0.21328 Kgm=0.21328\space Kg

m=0.21 Kg\colorbox{aqua}{m=0.21 Kg}


therefore mass of melted ice is 0.21 Kg in 2 minutes.






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