Anonymous
Anonymous asked in Education & ReferenceHomework Help · 2 months ago

# If 50.0ml of 40.0 degree celsius water is added to 200. ml of 90.0 degree Celsius water, what is the final temperature of the water?

What do I need to start I don't  get this🤦‍♂️🤦‍♂️🤦‍♂️🙏

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• m*s*t= 200*s*90=18000 cal

m*s*t= 50*s*40=2000 cal

s= specific heat of water=1

total vol 250 ml

total heat 20000 cal

temp =2000/250 = 80 deg C.

• touche

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• Let the final temperature of the mixture  = t°C

Heat given by 200 ml of water  =  Heat received by 50 ml water

If S = specific Heat of water,

=> 200 * S * ( 90 -  t )  =  50 * S * ( t - 40 )

=> 200 ( 90 - t )  =  50 ( t - 40 )

=> 4 ( 90 - t )  =  ( t - 40 )

=> 360 - 4 t   =   t - 40

=> 360 + 40   =  5 t

=>  5 t  =  400

=>  t  =  80° C  ............. Answer

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• 50(40) + 200(90) = 250x

2000 + 18000 = 250x

20000 = 250x

x = 20000/250

x = 80

80.0 ºC <––––––

• here we are again simple souls

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• I'm going to assume that somewhere amongst your reading material you were given a formula that deals with water temperatures.

Find it.

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