Adding Ice to Water An insulated beaker with negligible mass contains liquid wat

Question

Adding Ice to Water An insulated beaker with negligible mass contains liquid water with a mass of 0.305 kg and a temperature of 62.4 degree C Part A How much ice at a temperature of -20.4 degree C must be dropped into the water so that the final temperature of the system will be 27.0 degree C ? Take the specific heat of liquid water to be 4190 J/kg K the specific heat of ice to be 2100 J/kg .K ,and the heat of fusion for water to be 3.34 times10^5 J/kg. Not quite. Check through your calculations; you may have made a rounding error or used the wrong number of significant figures.

Explanation / Answer

given m = 0.305

temperatures are 62.4 , -20.4 , 27

the mass of the ice is say M

Qheat energy= m X C X dT

Qliquid= 0.305 X 4190 X (62.4 - 27 )

= 45239.43 J

the heat gained by ice is

Qice + Qfusion + Qliquid

Qice = M X 2100 X ( 0 - ( -20.4 ) )

= 42840 M

Qfusion = M X 3.34 X 105

Qliquid = M X 4190 X 27

= 113130 M

loss of heat = gain in heat

Qliquid = Qice + Qfusion + Qliquid

45239.43 = 42840 M + M X 3.34 X 105 + 113130 M

45239.43 = 489970 M

M = 45239.43 / 489970

M = 0.09233 Kg

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