91Ó°ÊÓ

Mass of cup = 200 g

Initial Temp = 65 ^oC

We have a 200 gram cup of boiling tea that we'd want to chill down to $65°\mathrm{C}$ before we drink it, by putting a mass m of ice (initially at -15 ^oC ) into the tea Given that the specific heat $\mathrm{cal}·{\mathrm{g}}^{-1}·{\mathrm{K}}^{-1}$. Assume that the tea have the same heat capacity as pure water 1 $\mathrm{cal}·{\mathrm{g}}^{-1}·{\mathrm{K}}^{-1}$, the tea must decrease by $35\mathrm{K}$so it must give up an amount of heat:

$Q_{\text{neater}}=mc\mathrm{Δ}T$

$Q_{\text{tea}}=200×1×\left(35\right)-7000\mathrm{cal}$

This heat goes into, first, heating the ice by 15 degrees to its melting point, then melting it, then heating the resulting water to 65 ^oC.

For the first step, the required heat raises the temperature of ice to its melting point is:

$Q_1=mc\mathrm{Δ}T$

where m is the mass of the ice, c is the specific heat of ice $0.5\mathrm{cal}·{\mathrm{g}}^{-1}·{\mathrm{K}}^{-1}$, and$∆T$ is the temperature difference between the initial temperature of ice and the melting point, so

role="math" localid="1650148419303" $Q_1=m×0.5×(0-(-15\left)\right)=7.5\mathrm{mcal}$

In the second step, the amount of heat required to melt the ice is:

$Q_2=m·L$

where L is the latent heat, and it's $80\mathrm{cal}/\mathrm{g}$for melting ice, so:

$Q_2=m·80=80\mathrm{mcal}$

- In the third step, the amount of heat lost by the tea to make the resulting water temperature at $65°\mathrm{C}$is,

$Q_3=mc\mathrm{Δ}T$

where m is the mass of the melted ice (water), c is the specific heat of water $1\mathrm{cal}·{\mathrm{g}}^{-1}·{\mathrm{K}}^{-1}$and $∆T$is the temperature difference between the initial temperature of melted ice and the final temperature of the mixture at $65°\mathrm{C}$, so:

$Q_3=m×1×(65-0)$

=65 mcal

The sum of these three heats is equal to the heat lost by the tea is:

$Q_{\mathrm{cas}}=Q_1+Q_2+Q_3$

$=7.5\mathrm{mcal}+80\mathrm{mcal}+65\mathrm{mcal}$

$=152.5\mathrm{mcal}$

Therefore:

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