This answer I tried before was wrong can someone help me? An insulated beaker wi
ID: 1472420 • Letter: T
Question
This answer I tried before was wrong can someone help me?
An insulated beaker with negligible mass contains liquid water with a mass of 0.305 kg and a temperature of 82.7 degree C . liquid water with a mass of 0.305 kg and a temperature of 82.7 degree C . How much ice at a temperature of -19.1 degree C must be dropped into the water so that the final temperature of the system will be 34.0 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 *10^5 J/kg .Explanation / Answer
By law of conservation of energy
Heat lost by water - heat gained by ice = 0
mwcw(Tf - Ti) - [ mici(Tf - Ti) + miLf + mici(Tf - Ti) ] = 0
mwcw(Tf - Ti) - mici(Tf - Ti) - miLf - mici(Tf - Ti) = 0
0.305*4190*(82.7 - 0) - mi*2100*(0 -(-19)) - mi*3.34*10^5 - mi*4190*(34-0) = 0
mi = 0.205 kg