# If 40 gram of water at 100 degree celsius is mixed with 150 gram of water and mixture temperature becomes 50 degree celcius. Find initial temperature of

cold water.

2
by Souvik1

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cold water.

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by Souvik1

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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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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

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