Question

500 g of water at 18 ° C and 0 g of 80 g of ice cubes are added. Dish own heat capacity is 100 J / K. Do you melt everything? If it melts, then what is the temperature of the final state, and if not melt, how much ice will remain without melting?

Answer 1

Check up if the heat lost by water and dish will be enough to melt all the ice.

Heat required to melt all the ice is

$q_1=m \Delta H_{fus}=80g\times334\frac{J}{g}=26720J$

Heat lost by water when temparature drops from$18^\circ C to 0^\circ C$ is

$q_2=cm \Delta T=4.186\frac{J}{g\times^\circ C}\times500g\times(0^\circ C -18^\circ C)= -37674 J$

 

Heat lost by dich is

$q_3=c\times \Delta T = -100\frac{J}{^\circ C}\times (0^0C-18^0C) = -1800 J$

Total lost of heat

$q_2 +q_3 = -37674J -1800J = -39474 J$

We can see that $q_1 <q_2 + q_3$  then all the ice will melt.

Determine the temperature of the final state:

$q_{gained} =q_{lost}$

$q_{gained} = q_1 +q_4$

$q_1=m \Delta H_{fus}=80g\times334\frac{J}{g}=26720J$

​

Heat gained by 80 g of melted ice when the temperature raises from $0^\circ C to T_{final}$  is

$q_4 = cm \Delta T$

$q_4 = 4.186f\frac{J}{g\times^\circ C}\times80g\times(T_{final} -0^\circ C)$

$q_{lost} = -(q_2+q_3) = -(4.186\frac{J}{g\times^\circ C}\times500g\times(T_{final} -18^\circ C) + 100\frac{J}{^\circ C} \times(T_{final} - 18^\circ C)$

$q_{lost} =q_{gained}$

$26720 + 4.186 \frac{J}{g\times^\circ C}\times80g\times(T_{final} -0^\circ C)= -(4.186 \frac{J}{g\times^\circ C}\times500g\times(T_{final} -18^\circ C) + 100\frac{J}{^\circ C}\times(T_{final} - 18^\circ C))$

$T_{final} = 5.05 ^\circ C$

Answer: $T_{final} = 5.05 ^\circ C$