A perfectly insulated calorimeter contains 500ml of water at30 o C and 45g of ic
ID: 1727135 • Letter: A
Question
A perfectly insulated calorimeter contains 500ml of water at30oC and 45g of ice at 0oC. Determine thefinal temperature of the system. A)21oCB)23.9oC
C)17.8oC
D)20.3oC
E)0oC, some ice remains. Hint: 1 ml = 1g therefore m(water) = 0.5 kg Energy conservation: The amount of heat lost by the water is equal to the amount ofheat absorbed by the ice as it melts into water and then getswarmer. DQ(water) = 0.5kg*(30 deg C-T)*1kcal/(kg deg C) = amount ofheat lost by the water as it cools to temperature T(in degC). DQ(ice) = 0.045kg * 80 kcal/kg + 0.045kg*(T-0 deg C))*1kcal/(kg deg C) = amount of heat absorbed by ice as it melts andwarms up. Set DQ(water) equal to DQ(ice) and solve for T, the finaltemperature.
A perfectly insulated calorimeter contains 500ml of water at30oC and 45g of ice at 0oC. Determine thefinal temperature of the system. A)21oC
B)23.9oC
C)17.8oC
D)20.3oC
E)0oC, some ice remains. Hint: 1 ml = 1g therefore m(water) = 0.5 kg Energy conservation: The amount of heat lost by the water is equal to the amount ofheat absorbed by the ice as it melts into water and then getswarmer. DQ(water) = 0.5kg*(30 deg C-T)*1kcal/(kg deg C) = amount ofheat lost by the water as it cools to temperature T(in degC). DQ(ice) = 0.045kg * 80 kcal/kg + 0.045kg*(T-0 deg C))*1kcal/(kg deg C) = amount of heat absorbed by ice as it melts andwarms up. Set DQ(water) equal to DQ(ice) and solve for T, the finaltemperature.
Hint: 1 ml = 1g therefore m(water) = 0.5 kg Energy conservation: The amount of heat lost by the water is equal to the amount ofheat absorbed by the ice as it melts into water and then getswarmer. DQ(water) = 0.5kg*(30 deg C-T)*1kcal/(kg deg C) = amount ofheat lost by the water as it cools to temperature T(in degC). DQ(ice) = 0.045kg * 80 kcal/kg + 0.045kg*(T-0 deg C))*1kcal/(kg deg C) = amount of heat absorbed by ice as it melts andwarms up. Set DQ(water) equal to DQ(ice) and solve for T, the finaltemperature.
Explanation / Answer
Given that the mass of water is m = 500 ml = 0.5kg Initial temperature is T1 = 30oC Mass of ice is M = 0.045 kg Initial temperature of ice is T2 = 0oC ------------------------------------------------------------------- Heat required to melt the ice at 0oC is Q1 =M*Lice = 0.045kg*333000 J/kg = 14985 J Heat given by the water change from T1 to0oC Q2 = m*s*T1 = 0.5kg*4186 J/kg.K*30oC = 62790 J Q2 > Q1 so the ice completely melt Now we find final temperature of system as Heat lost by water = heat gained by ice m*s*(T1 - T) = M*Lice + M*s*T T = ( m*s*T1 - M*Lice ) / ( M*s + m*s) = 47805 / 2281.4 =20.95oC where specific heat of water is s = 4186 J/kg.K Latent heat of ice is Lice = 333000 J/kg
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