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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Thermal Properties of Matter

NTA Abhyas 2020 · Physics · Thermal Properties of Matter

One end of a copper rod of uniform cross-section and of length 1....

One end of a copper rod of uniform cross-section and of length 1.5m1. 5m is kept in contact with ice at 0°C0 ° C and the other end with water at 100°C100 ° C. If the whole system is insulated from its surroundings,& then at what& distance from the end of the rod which is in contact with the steam, the temperature& should be& maintained at 200°C200 ° C, so that in steady-state the rate of melting of ice be equal to the rate of production of steam? [Lfusion=80calg−1[ {L}_{fusion} = 80calg^{- 1}, Lvaporization=540calg−1]{L}_{vaporization} = 540calg^{- 1} ]

One end of a copper rod of uniform cross-section and of length 1. 5m is kept in contact with ice at 0 ° C and the other

Options

  1. A10.34cm10. 34cm
  2. B16.48cm16. 48cm
  3. C21.68cm21. 68cm
  4. D27.87cm27. 87cm

Correct answer

A. 10.34cm10. 34cm

Step-by-step solution

If the point is at a distance x from water at 100oC, heat conducted to ice in time t,

Qice=KA(200−0)(1.5−x)×t{Q}_{ice} = \text{KA} \dfrac{\left(\text{200} - 0\right)}{\left(\text{1.5} - x\right)} \times \text{t}

One end of a copper rod of uniform cross-section and of length 1. 5m is kept in contact with ice at 0 ° C and the other

So ice melted by this heat

mice=QiceLF=KA80(200−0)(1.5−x)×t{m}_{ice} = \dfrac{{Q}_{ice}}{{L}_{F}} = \dfrac{\text{KA}}{\text{80}} \dfrac{\left(\text{200} - 0\right)}{\left(\text{1.5} - x\right)} \times \text{t}

Similarly heat conducted by the rod to the water at 100oC in time t,

Qwater=KA(200−100)xt{Q}_{water} = \text{KA} \dfrac{\left(\text{200} - \text{100}\right)}{x} \text{t}

So steam formed by this heat

msteam=QwaterLV=KA(200−100)540×xt{m}_{steam} = \dfrac{{Q}_{water}}{{L}_{V}} = \text{KA} \dfrac{\left(\text{200} - \text{100}\right)}{\text{540} \times x} \text{t}

According to given problem mice = msteam, i.e.,

20080(1.5−x)=100540×xorx=658m=10.34 cm\dfrac{\text{200}}{\text{80} \left(\text{1.5} - x\right)} = \dfrac{\text{100}}{\text{540} \times x} \text{or} x = \dfrac{6}{\text{58}} \text{m} = \text{10.34 cm}

i.e., 200oC temperature must be maintained at a distance 10.34 cm from water at 100oC.
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