Here in this problem there are three stages.

One is the energy required to melt ice.

Second one to rise the temperature of melted ice from0 0 C to the final temperatureT f .

And third one is the energy lost by the water to from thetemperature 80 0 C to the finaltemperature(T f ). Now according to the law of conservation of energy the amountof heat taken is equal to the amoount of heat lost.

So \(\displaystyle{m}_{{{i}{c}{e}}}{L}_{{{f}}}+{m}_{{{i}{c}{e}}}{C}_{{{w}{a}{t}{e}{r}}}{\left({T}_{{{f}}}-{0}^{{{0}}}{C}\right)}={m}_{{{w}{a}{t}{e}{r}}}{C}_{{{w}{a}{t}{e}{r}}}{\left({80}^{{{0}}}-{T}_{{{f}}}\right)}\)

Here c water = 4186J/kg. 0

C

L

\(\displaystyle{f}={3.33}\times{10}^{{{5}}}\) J/kg And m ice = 100g = 0.1kg

mwater =1.0kg

Now substitute all the given values in the above mentionedformula and then solve for the value of the finaltemperature(Tf).

One is the energy required to melt ice.

Second one to rise the temperature of melted ice from0 0 C to the final temperatureT f .

And third one is the energy lost by the water to from thetemperature 80 0 C to the finaltemperature(T f ). Now according to the law of conservation of energy the amountof heat taken is equal to the amoount of heat lost.

So \(\displaystyle{m}_{{{i}{c}{e}}}{L}_{{{f}}}+{m}_{{{i}{c}{e}}}{C}_{{{w}{a}{t}{e}{r}}}{\left({T}_{{{f}}}-{0}^{{{0}}}{C}\right)}={m}_{{{w}{a}{t}{e}{r}}}{C}_{{{w}{a}{t}{e}{r}}}{\left({80}^{{{0}}}-{T}_{{{f}}}\right)}\)

Here c water = 4186J/kg. 0

C

L

\(\displaystyle{f}={3.33}\times{10}^{{{5}}}\) J/kg And m ice = 100g = 0.1kg

mwater =1.0kg

Now substitute all the given values in the above mentionedformula and then solve for the value of the finaltemperature(Tf).