Question

can someone tell me how they got this answer

0/1pts Question 7 Ice with a mass of 123,588 kg at-10.5°C is used to cool a beluga whale's water tank, which holds 1,017 m3 of water. If the water tank was initially at 19.0°C. to what final equilibrium temperature is the tank brought to? Answered margin of error +-2 % ect Answer 7.7535 Consider the following setup: Qice +Qmelt+ Qice-water +Qater = 0 Where Qice is the amount of thermal energy necessary to warm the ice to 0° C Qice =mice Cice (0° C--10° C) Qmelt is the amount of energy necessary to completely melt the ice. Qmelt = mice Lf ice Qice oater is the amount of thermal energy required to warm the added mass of the original ice from 0° C to the final temperature (you must solve for the final temperature). mice Cwater (T-0° C) Qice-water e e

Answer 1

Note : heat lost from water which is at higher temp. This heat is gain by ice at -10.5 ℃... After heat gain, ice at -10.5℃ raises its temp upto 0℃. in this state, ice is still solid. after reaching at 0℃, ice melts and convert into 0℃ water. now this ice water temp again raises to final equilibrium temp Tf.

water which is at 19℃ coverts into water at Tf temp.

Qwiesen Question > oc rõ at į Ice water d alors'c y water Ice water/ Considering, heat troms for occurs bio water ice & water only TF at & NO cheat loss to 19°C Sworounding Quotes - Hence, in thermal equiwibriumaa : ice t a matt Rice water + Twater - o Cuader* (TF_o) Micea Cice r on (-1015)] & Mice*lf & Mice-coater sri) + Muotes Cwoter * (76-19)=0 & here micé 123588 KE - mice. Water Lf a latent heap of fusion of ice = 334 kolla Sice = Spocheat of ice = 2.108 kg kg-c . Cwater a sp. heat of water a 4.187 kg 11g="C Tf = final equilibrium timp. of Ice & water . srstem mwaters Swim voda = 1000x1017 = 1017000 kg I from egr cis} 423588* 52.108 * 105 + 334+ 4-187* 123588* Tf + 1017000 x4-187*(15-19) 4775641-956* Tf - 36891512.21 - If = 7072499c 4ms TfE7'7249°C