A $$10$$ W electric heater is used to heat a container filled with $$0.5$$ kg of water. It is found that the temperature of water and container rises by $$3K$$ in $$15$$ min. The container is then emptied, dried and filled with $$2$$ kg of oil. The same heater now raises the temperature of container oil system by $$2K$$ in $$20$$ min. Assume there is no heat loss in the process and the specific heat of water is $$4200$$ J $$kg-1K-1$$ , the specific heat of oil in the same limit is equal to

Question

A $$10$$ W electric heater is used to heat a container filled with $$0.5$$ kg of water. It is found that the temperature of water and container rises by $$3K$$ in $$15$$ min. The container is then emptied, dried and filled with $$2$$ kg of oil. The same heater now raises the temperature of container oil system by $$2K$$ in  $$20$$ min. Assume there is no heat loss in the process and the specific heat of water is $$4200$$ J $$kg-1K-1$$ , the specific heat of oil in the same limit is equal to

Infinity Learn · Accepted Answer

As we know, work done by an electric heater i.e.  $$m1c1$$Δ$$t+m2c2$$Δ$$t=$$ work done So,    $$m1c1$$Δ$$t+m2c2$$Δ$$t=P1t1where$$,  $$m1=0.5kg$$, specific heat, $$c1=4200Jkg$$−$$1K$$−$$1$$Δ$$t=$$Δ$$t1=$$Δ$$t2=3KP1=10Wt1=15$$×$$60=900sc2$$ $$=$$ specific heat capacity of container. So, $$0.5$$×$$4200$$×$$(3$$−$$0)+m2c2$$×$$(3$$−$$0)=10$$×$$15$$×$$60$$⇒ $$2100$$×$$3+m2c2$$×$$3=9000$$⇒ $$m2c2=9000$$−$$63003=900Similarly$$, in the case of oil,  $$m1co$$Δ$$t+m2c2$$Δ$$t=P2t2$$ where, $$co=$$ specific heat capacity of oil, $$P2=10W2$$×$$c0$$×$$2+900$$×$$2=10$$×$$20$$×$$60$$⇒ $$4c0+1800=12000$$⇒ $$co=2550=2.55$$×$$103Jkg$$−$$1K$$−$$1$$

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