91Ó°ÊÓ

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 $1\mathrm{cal}â‹\dots {\mathrm{g}}^{−1}â‹\dots {\mathrm{K}}^{−1}$, the tea's heat capacity must fall by ${35}^{∘}$, implying that it must give up some heat:

$Q_{\text{water}}=mc\mathrm{Δ}T$

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

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

$Q_1=mc\mathrm{Δ}T$

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

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

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

$Q_2=m×L$

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

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

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

$Q_3=mc\mathrm{Δ}T$

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

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,

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

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

$→m=45.9\mathrm{g}$