Answers

2014-09-05T15:50:56+05:30

we have a formula T=(mt1+m2t2)/(m1+m2)

here m1=mass of water=40g

         t1=temperature of water=100deg c

          m2=  mass of second water taken=150

          t2=temperature of second water taken

          T=50

50=40(100)+150(t2)/(40+150)

50=(4000+150t2)/190

9500=4000+150t2

9500-4000=150t2

5500=150t2

5500/150=t2 

36.6=t2

therefore the intial temperature of cold water is 36.6deg celcius


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2014-09-07T15:34:00+05:30

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Let s be the specific heat capacity of water. Let the temperature of cold water be T deg Celsius.

Heat lost by hot water = heat gained by cold water
m1 s ΔT = m2 s ΔT

40gm  * s Cal/degK /gm * (100 - 50) degK = 150gm * s cal/gm/degK * (50 - T)
40 * 50 = 150 (50-T)
2000 = 7500 - 150 T

150 T = 5500

T = 36.666 deg Celsius


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