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A thin metallic wire of resistance `100 Omega` is immersed in a calorimeter containing 250 g of water at `10^@C` and a current of 0.5 ampere is passed through it for half an hour. If the water equivalent of the calorimeter is 10 kg, find the final temperature of water. |
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Answer» Here, `R = 100 Omega, m = 250g`, `theta_(1) = 10^@C, I = 0.5 A, t = 30 xx 60 s, w= 10g`, `theta_(2)` = final temperature of water (in `.^(@)C`) Heat produced, `H = I^(2) Rt` `= (0.5)^(2) xx 100 xx (30 xx 60)= 45000 J` Heat gained by water and caloimeter `= (m +w) c(theta_(2)- theta_(1))` `=(250 + 10) xx 1 xx (theta_(2) - theta_(1)) cal` `= 260 xx 4.2 xx (theta_(2)- theta_(1))= J` `:. 260 xx 4.2 xx (theta_(2)- theta_(1))=45000` or `theta_(2) - theta_(1) = 45000/(260 xx 4.2) = 41.2^@C` `theta_(2) = 41.2 + theta_(1) = 41.2 + 10 = 51.2^(@)C` |
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