Physics - Elasticity Question with Solution | TestHub

To find the final temperature, we calculate the heat required to melt the ice and the heat released by the water.

Given:

Mass of ice ($m_i$) = 540 g

Initial temperature of ice ($T_i$) = $0^{\circ }\mathrm{C}$

Mass of water ($m_w$) = 540 g

Initial temperature of water ($T_w$) = $80^{\circ }\mathrm{C}$

Latent heat of fusion of ice ($L_f$) = 336 J/g

Specific heat capacity of water ($c_w$) = 4.2 J/g${}^{\circ }\mathrm{C}$

Heat absorbed by ice to melt at $0^{\circ }\mathrm{C}$:

$Q_1=m_i{L}_f=540g\times 336J/g=181440J$

Heat released by water to cool from $80^{\circ }\mathrm{C}$ to $0^{\circ }\mathrm{C}$:

$Q_2=m_w{c}_w\Delta T=540g\times 4.2J/g^{\circ }\mathrm{C}\times (80^{\circ }\mathrm{C}-0^{\circ }\mathrm{C})$

$Q_2=540\times 4.2\times 80=181440J$

Since the heat required to melt all the ice ($Q_1$) is equal to the maximum heat that can be released by the water to cool to $0^{\circ }\mathrm{C}$ ($Q_2$), all the ice melts, and the final temperature of the mixture is $0^{\circ }\mathrm{C}$.

The final temperature of the mixture is $0^{\circ }\mathrm{C}$.