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10 g of water at 80℃ is mixed with 50g of copper at 10℃. Find the final temperature |
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Answer» Explanation:since it is given that 50 gm water at 20 ⁰ C and 50 gm of water at 40⁰ C are mixed. Since the masses of the liquid at different TEMPERATURES are same, the answer is very easy and simple : average of 20⁰C and 40⁰C. that is: 30⁰C.final temperature of the mixture = [ m 1 * T 1 + m 2 * T 2 ] / (m 1 + m 2)= [ 50 gms * 20⁰ C + 50 gms * 40⁰C ] / (50+50) = 3,000 / 100 = 30⁰C another way using specific heats : LET the final temperature be = T ⁰C AMOUNT of heat given out by the hot water = m * s * (40⁰C - T) = 50 gms * s* (40 -T) Amount of heat taken in by the cold water = m * s * (T - 20⁰C) = 50 gms * s * (T - 20 ) As the amounts are equal, because the heat is transferred from hotwater to the cold water without any loss of heat to any surroundings, 50 * s * (40 -T) = 50 gm * s * (T-20) 40 - T = T - 20 2 T = 60 => T = 30⁰C |
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