For a copper-iron $(Cu-Fe)$ couple,the thermo $e.m.f.$ (temperature of cold junction $= 0\,^{\circ}C$) is given by $E = (14\theta - 0.02\theta^2)\,\mu V$. The neutral temperature will be

  • A
    $350\,^{\circ}C$
  • B
    $350\,K$
  • C
    $560\,^{\circ}C$
  • D
    $560\,K$

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Similar Questions

Column $I$ gives some devices and Column $II$ gives some processes on which the functioning of these devices depends. Match the devices in Column $I$ with the processes in Column $II$.
Column $I$Column $II$
$(A)$ Bimetallic strip$(p)$ Radiation from a hot body
$(B)$ Steam engine$(q)$ Energy conversion
$(C)$ Incandescent lamp$(r)$ Melting
$(D)$ Electric fuse$(s)$ Thermal expansion of solids

How much energy will be released by cooling $1\,kg$ of water up to a temperature of $10\,^{\circ}C$?

Two different metal wires are joined to form a thermocouple. One junction is kept at a constant temperature,while the other junction is heated to a higher temperature. Which graph represents the variation of thermo $e.m.f.$ $(E)$ with temperature $(t)$?

$A$ lead bullet at $27^{\circ}C$ just melts when stopped by an obstacle. Assuming that $25\%$ of the heat produced is absorbed by the obstacle,find the velocity of the bullet at the time of striking in $m/s$. (Melting point of lead $= 327^{\circ}C$,specific heat of lead $= 0.03 \, cal/g^{\circ}C$,latent heat of fusion of lead $= 6 \, cal/g$,and $J = 4.2 \, J/cal$)

$A$ refrigerator converts $500\,g$ of water at $25\,^{\circ}C$ into ice at $-10\,^{\circ}C$ in $3\,hours\,40\,minutes$. The quantity of heat removed per minute is ........ $cal/\min$.
(Specific heat of water $= 1\,cal/g\,^{\circ}C$,Specific heat of ice $= 0.5\,cal/g\,^{\circ}C$,Latent heat of fusion $= 80\,cal/g$)

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