Three containers C1,C2 and C3 have water at different temperatures. The table below shows the final temperature T when different amounts of water (given in liters) are taken from each container and mixed (assume no loss of heat during the process)The value of θ (in 0C to the nearest integer) is -----------
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Detailed Solution
First Mixing : 1s θ1−60+2sθ2−60+0sθ3−60=0⇒θ1+2θ2=180 …………. 1 Second Mixing : 0s θ1−30+1sθ2−30+2sθ3−30=0⇒θ2+2θ3=90 …………. 2 Third mixing : 2s θ1−60+0sθ2−60+1sθ3−60=0⇒2θ1+θ3=180 …………. 3 Solving equations 1 2 and 3 we get θ1=80, θ2=50, θ3=20 Fourth mixing : 1s θ1−θ+1sθ2−θ+1sθ3−θ=0 ⇒80−θ+50−θ+20−θ=0 ⇒θ=1503=50