Question

Some ice cubes at 0°C with a total mass of 441 g are placed in a microwave oven and subjected to 750. W (750. J/s) of energy for 5.00 minutes. (Assume all the energy of the microwave is absorbed by the water and no heat loss by the water.) (Enter your answer to three significant figures.) What is the final temperature of the water? C

Answer 1

First Calculate the value of total amount of energy added to the ice

Energy = Power * time

=5 min x 60 s/min x 750 J/s = 225,000 J

we know that the ice cubes are at 0 deg C already,hence first we need to Calculate the heat required to melt the ice at 0 C

Heat required = mass in g x heat of fusion of water in J/g

= 441 g x 334 J/g

= 147,294 J

Now we substract this much of energy which is needed to convert all the ice to water from the value of total energy total, remaining amount of energy will increase the temperature ofwater

= 225000 - 147294 = 77,706 J

Then applying the formula of heat capacity of liquid water

delta H = m * C * (T - To)

C= specific heat capacity of water

C for liquid water is 4.18 J/gK

therefore the final temp T = H/(m*C) + To, where To = 0

so T = 77,706 J/(441g * 4.18 J/gK) = 42.1 degrees

PLEASE LEAVE THE POSITIVE FEEDBACK IF YOU LIKE THE SOLUTION