this is the question If 7.30 kg of ice at -5.00°C is added to 12.0 kg of water a

Question

this is the question

If 7.30 kg of ice at -5.00°C is added to 12.0 kg of water at 15°C, compute the final temperature. (which is t=0) and compute the amount of ice that remains

I looked at previous questions of this type but i get the wrong answer
this is what i did

Latent heat of fusion of water ( L ) = 3.33 *105 J/kg
Qmelt = mice L
= ( 7.30 kg ) * ( 3.33 * 10^5 J/kg ) = 2430900 J

Q water= m water c?T
= ( 12.0 kg ) * ( 4190 J/kg.o C ) ( -5oC -15oC )
= -1005600J
T = 0oC
Q water= m L
-1005600 =m ( 3.33 * 105 J/kg )
= 3.0198 kg

Thanks for helping!

Explanation / Answer

Heat absorbed by ice

mi * ci * (T0 - Ti) + (mi - m) * L = mw * cw * (Tw - T0)

specific heat of ice ci = 2.05 * 103  J/kg-0C

latent heat of fusion of ice L = 3.34 * 105  J/kg

specific heat of water cw   = 4.18 * 103  J/kg-0C

7.30 * 2.05 * 103 * {0 - ( - 5)} + (7.30 - m) * 3.34 * 105   = 12.0 * 4.18 * 103 * (15 - 0)

(7.30 - m) = (7.524 * 105   - 7.483 * 104) / 3.34 * 105

= 2.03

mass of ice that remains m = 7.30 - 2.03

= 5.27 kg

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