promotion image of download ymail app
Promoted
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?

Pleaseeee help me 

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

4 Answers

Relevance
  • 2 months ago

    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.

  • 2 months ago

    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

    • Commenter avatarLogin to reply the answers
  • 2 months ago

    50(40) + 200(90) = 250x

    2000 + 18000 = 250x

    20000 = 250x

    x = 20000/250

    x = 80

    80.0 ºC <––––––

  • VP
    Lv 7
    2 months ago

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

    Find it.  

        Reread it. 

            Implement it.

    • charlatan
      Lv 7
      2 months agoReport

      help me ,there is no help here

    • Commenter avatarLogin to reply the answers
Still have questions? Get your answers by asking now.