Question

an ice cube at 0.00 degrees celsius with a mass of 4.52 g is placed into 55g of water, initially at 23 degrees celsius. If no heat is lost to the surroundings, what is the final temperature of the entire water sample after all the ice is melted? (Specific heat of water is 4.184J/g*degrees celsius)

Answer 1

Dear student,

Finstly we have take into account phase change underwent by the solid walen at o'n to liquid at orc Heat needed to melt the solid at its melting point will come from the warmen water sample qutqa a -43 Whenei 9,- the head absorbed by the solid at 0°C 2 = the heat absorbed by the liquid at o'c 9/= the heat lost by the warmer water sample We know that q=m.. AT where, q=heat absorbed on losh m= mass of sample c = the specific cheat of water The amount of heat required by the ice do convent to weden (i= letent heal fusionofice! = 333.63 lgm

OF 4.52 garx 33 3:6 Igor = 1.5x 103 J Bo 1.5RJ + 92 = -93 The minus is just to show the heat lost. Va = mc (Alice) = 46 52 gun (4.25 / 4) (74-8) * = (4.52 x 4.2) (T4-0)J (specific heat of waten = 4:1851gc = 18.98 (+4 -0.)] ~ 4.25/6) Az = mc (DT water) = 55gr 14.25 / gist) (23 - The = (55x4-2) (23-14) • 231 (23-T;)J 9 3 = 91 +92 231 (23-1f) 7 = 1.3 x 103 533 - 231Tf = 1500 + a 18.98 (Tf-0) * - 18.98 + 18.98 If

(5313 -1481) = 231 If + 18.9877 3832°C = 249.98 Tf = 3832 . 249.98 Jo Tf = 15:32: The final temperature of mixture is 15.08°C

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