aspx lass Management 1 Help mework 14 Begin Date: 4/4/2016 12:00:00 AM-Due Date:

Question

aspx lass Management 1 Help mework 14 Begin Date: 4/4/2016 12:00:00 AM-Due Date: 4/18/2016 11:00:00 PM End Date: 5/2/2016 11:00:00 PM (10%) Problem 4: Soda from a ms = 12 oz can at temperature rss/95" C is poured in its entirety into a glass containing a mass m, . 0.13 kg amount of ice at temperature T,--155. C. Assume that ice and water have the following specific heats: q-2090 J/(kg C) and es 4186J(kg*C), and the latent heat of fusion of ice is Ly 334 kikg. 33% Part (a) In degrees celsius, what is the final temperature T,inal of the mixture? 33% Part (b) Write an expression for how much of the ice mi," has melted? Grade : Deducti Potentia Submis Attemp 123 1 2 Submit Hint I give up? deduction per feodback Hints,1% doartin per hint. Hints remaining Feedback: 0 33% Part (c) In kilograms, how much of the ice mm has melted? Au anun.c2016 ExpettA.LLC

Explanation / Answer

ms = 12 oz = 340.2 gm
Ts = 19.5 oC
mi = 0.13 Kg = 130 gm
Ti = -15.5 oC

Specific Heat of Soda, Cs = 4.186 J/gm k
Specific Heat of ice, Ci = 2.090 J/gm k
Latent Heat of ice, Li = 334.0 J/gm

(a)

Heat Gained by Ice = Heat Lost by soda

Let us check,
Heat needed to completely melt ice,
Q = mi * Ci * T + mi * Li
Q = 130 * 2.090 * 15.5 + 130 * 334 J
Q = 47631.35 J

Now,
Heat Given by Soda if we bring it to 0o C,
Q =  ms * Cs * T
Q = 340.2 * 4.186 * 19.5
Q = 27769.5 J

So we can clearly see, Heat given by soda is not sufficient to completely melt the ice.
Therefore Final Temperature of the mixture, Tfinal = 0o C

(b)
Expression for how much of ice has melted,
Let mm grams of ice have melted then we can write -
mm * Li + mi * Ci * T = ms * Cs * T

(c)
Substituing Values in above eq,
mm * 334 + 130 *  2.090 * 15.5 = 27769.5
mm = 70.53 gm
mm = 0.705 Kg

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