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A 10 watt electric heater is used to heat a container filled with 0.5 kg of water. It is found that the temperature of water and the container rises by 3^@ K in 15 minutes. 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 2^@ K in 20 minutes. Assuming that there is no heat loss in the process and the specific heat of water as 4200 J kg^(-1)K^(-1)the specific heat of oil in the same unit is equal to |
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Answer» `1.50 xx 10^3 ` Massofwater`m_w= 0.5Kg` Massof oil`m_0=2kg ` specificheatof water`s_w= 4200J Kg^(-1) K^(-1)` incase CONTAINER watersystem energysuppliedbythe heaterto thesystemin 15min `= 10 xx 15 xx 60= 90 00 J` ` thereforem_w s_w DeltaT_w+m _c S_cDeltaT_c= 9000 ` ( wheresubscriptsw andc refereto watereand continerrespectively ). `0.5 xx 4200 xx 3 xx m_cxxs_cxx 3=9000` ` 6300+ 3m_cs_c=9000or3m_C s_c2700` `m_c s_c = 900` in caseof container-oilsystem energysupplied by thesame heaterto the systemin 20min `=10 xx 20 xx 60 = 1200 J` ` thereforem_o S_0DeltaT_o+ m_cS_cDeltaT_c= 12000` ( wheresubjscripto referto oil ) ` 2 xx s _0xx2+ 900xx 2= 1200 ` ` 2 xx S_0xx 2+ 900xx 2= 1200 ` ` 4s_0=10200` ` s_0 =2550j Kg^(-1)K^(-1) =2.55 xx 10^3J kg^(-1) K^(-1)` hencethe specificheatof OILIS `2.55 xx 10^3Jkg^(-1)K^(-1)` |
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