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प्रश्न
One end of a steel rod (K = 46 J s−1 m−1°C−1) of length 1.0 m is kept in ice at 0°C and the other end is kept in boiling water at 100°C. The area of cross section of the rod is 0.04 cm2. Assuming no heat loss to the atmosphere, find the mass of the ice melting per second. Latent heat of fusion of ice = 3.36 × 105 J kg−1.
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उत्तर
Given:
Thermal conductivity of the rod, K = 46 J s–1 m–1 °C–1
Length of the rod, l = 1 m
Area of the cross-section of the rod, A =0.04 cm2
= 0.04 × 10−4 m2
= 4 × 10−6 m2
Rate of transfer of heat is given by
`(ΔQ)/(Δt) = (ΔT)/ (l/(KA)`
`(ΔQ)/(Δt) = ((T_1 - T_2 ) KA ) / l`
`(ΔQ)/(Δt) =( (100-0 )/1) xx 46 xx 4 xx 10^-6 m^2`
`(DeltaQ)/(Deltat) = 184xx10^-4 J//s`
Also , ΔQ = mL f
`therefore (m/t)L_f = 184 xx 10^-4`
`⇒ ( m/t) xx 3.36 xx 10^5 = 184 xx 10^`
`rArr m=(184xx10^-4)/(3.36xx10^5)xxt`
`rArr m = 5.5xx10^-9xx1 kg//s`
`⇒ m = 5.5 xx 10^-8 xx 10^3 g/s`
`⇒ m = 5.5 xx 10^-5 g/s`
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