This answer I tried before was wrong can someone help me? An insulated beaker wi

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

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