Question

Your $200g$cup of tea is boiling-hot. About how much ice should you add to bring it down to a comfortable sipping temperature of ${65}^{\circ }\mathrm{C}$? (Assume that the ice is initially at $-{15}^{\circ }\mathrm{C}$. The specific heat capacity of ice is $(\left.0.5\mathrm{cal}/\mathrm{g}{\cdot }^{\circ }\mathrm{C}.\right)$.)

Short answer

The sum of these three heat is equal to the heat lost by the tea$m=45.9\mathrm{g}$

Step-by-step solution

Step : 1 Thermal capacity

We have a $200$gram cup of boiling tea and want to chill it down to ${65}^{o}C$before drinking it, so we put a mass $m$of ice (at $-{15}^{O}C$) into it. Given that ice has a thermal capacity of Assuming that the tea has the same heat capacity as pure water $1\mathrm{cal}\cdot {\mathrm{g}}^{-1}\cdot {\mathrm{K}}^{-1}$, the tea's heat capacity must fall by ${35}^{\circ }$, implying that it must give up some heat:

$Q_{\text{water}}=mc\mathrm{\Delta }T$

$$\begin{gathered} Q_{tea}=200\times 1\times \left(35\right) \\ =7000\mathrm{cal} \end{gathered}$$

Step : 2  Stages of melting ice 

The required heat to raise the temperature of ice to its melting point in the first phase is:

$Q_1=mc\mathrm{\Delta }T$

where $m$is the ice's mass, $c$is the ice's specific heat $0.5\mathrm{cal}\cdot {\mathrm{g}}^{-1}\cdot {\mathrm{K}}^{-1}$is the temperature difference between the ice's original temperature and the melting point, so:

$$\begin{gathered} Q_1=m0.5\times (0-(-15\left)\right) \\ =7.5m\mathrm{cal} \end{gathered}$$

The amount of heat necessary to melt the ice in the second stage is:

$Q_2=m\times L$

where $L$is the heat $80\mathrm{cal}/\mathrm{g}$for melting ice

$$\begin{gathered} Q_2=m\times 80 \\ =80m\mathrm{cal} \end{gathered}$$

The amount of heat lost by the tea in the third phase to achieve a water temperature of ${65}^{0}C$is,

$Q_3=mc\mathrm{\Delta }T$

where $m$is the mass of melted ice (water), and $c$is water's specific heat.$1\mathrm{cal}\cdot {\mathrm{g}}^{-1}\cdot {\mathrm{K}}^{-1}\text{and}\mathrm{\Delta }T$ is the temperature difference between the beginning temperature of melting ice and the ultimate temperature of the combination at ${65}^{0}C$.

Step :3 Sum of three heat

Because the sum of these three heats equals the heat lost by the tea, equations $\left(1\right),\left(2\right),\left(3\right)and\left(4\right)$ can be used,

$Q_{tea}=Q_1+Q_2+Q_3$

$$\begin{gathered} 7000=7.5m+80m+65m \\ =152.5m \end{gathered}$$

$\to m=45.9\mathrm{g}$