Question

a) Two 50 g ice cubes are dropped into 200 g of water in a thermally insulated container. if the water is initially at 25 C, and the ice comes directly from a freezer at -15 c ,what is the final temperature at thermal equilibrium?
b)what is the final temperature if only one ice cube is used?

Answer 1

a.

Let us write down conservation energy condition. On the left side is heat given to ice cubes, on the right - heat taken from water

$2\cdot m_{ice}\cdot c_{ice}\cdot\Delta T_{0}+2\cdot m_{ice}\cdot\lambda+2\cdot m_{ice}\cdot c_{water}\cdot t_{1}=m_{water}\cdot c_{water}(25-t_{1})$

here $c$ denotes specific heat capacity and lambda denotes specific heat of melting of ice, $\Delta T_{0}=15{}^{\circ}$ C. We can find $t_{1}$ - the final temperature from this equation.

$t_{1}=\frac{m_{water}\cdot c_{water}\cdot 25-2\cdot m_{ice}\cdot c_{ice}\cdot\Delta T_{0}-2\cdot m_{ice}\cdot\lambda}{m_{water}\cdot c_{water}+2\cdot m_{ice}\cdot c_{water}}$

But if we evaluate

$m_{water}\cdot c_{water}\cdot 25-2\cdot m_{ice}\cdot c_{ice}\cdot\Delta T_{0}-2\cdot m_{ice}\cdot\lambda=0.2\cdot 4200\cdot 25-2\cdot 0.05\cdot 15\cdot 2060-2\cdot 0.05\cdot 335000=-15500$

we will see it is negative. this means that there is not enough heat to melt all the ice. Hence, final temperature is 0 C.

b.

The same formula without factor 2 for ice

$t_{1}=(m_{water}\cdot c_{water}\cdot 25-m_{ice}\cdot c_{ice}\cdot\Delta T_{0}degrees-m_{ice}\cdot\lambda)/(m_{water}\cdot c_{water}+m_{ice}\cdot c_{water})=$

$=\frac{0.2\cdot 4200\cdot 25-0.05\cdot 15\cdot 2060-0.05\cdot 335000}{0.2\cdot 4200+0.05\cdot 4200}=2.5{}^{\circ}C$

Here answer is 2.5 degrees C