A copper ring has a diameter of exactly $$25$$ mm at it’s temperature of $$0$$°C. An aluminum sphere has a diameter of exactly $$25.05$$ mm at its temperature of $$100$$°C. The sphere is placed on top of the ring and two are allowed to come to thermal equilibrium, no heat being lost to the surrounding. The sphere just passes through the ring at the equilibrium temperature. Find the ratio $$10mspheremring$$ $$(approximately)Given$$ α$$Cu=1.7$$×$$10$$−$$5/$$°C, α$$Al=2.3$$×$$10$$−$$5/$$°CSpecific heat of $$Cu=0.0923$$ $$cal/g$$°CSpecific heat of $$Al=0.215$$ $$cal/g$$°CSymbols have their usual meanings

Question

A copper ring has a diameter of exactly $$25$$ mm at it’s temperature of  $$0$$°C. An aluminum sphere has a diameter of exactly $$25.05$$ mm at its temperature of  $$100$$°C. The sphere is placed on top of the ring and two are allowed to come to thermal equilibrium, no heat being lost to the surrounding. The sphere just passes through the ring at the equilibrium temperature. Find the ratio $$10mspheremring$$ $$(approximately)Given$$   α$$Cu=1.7$$×$$10$$−$$5/$$°C,  α$$Al=2.3$$×$$10$$−$$5/$$°CSpecific heat of  $$Cu=0.0923$$ $$cal/g$$°CSpecific heat of  $$Al=0.215$$ $$cal/g$$°CSymbols have their usual meanings

Infinity Learn · Accepted Answer

In equilibrium, both have samefinal  lengthsfinal length, $$l2=l11+$$α$$Tf-Ti251+17$$×$$10$$−$$6(T-0)=25.051-2.3$$×$$10$$−$$5(100-T)$$⇒$$T=50$$∘Cmsphere ×$$0.215$$×(100$$−$$50)=mRing$$ ×$$0.0923$$×$$(50$$−$$0)⇒$$10msphere$$ mring $$=4$$

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