Question

Iced tea is made by adding ice to 1.8 kg of hot tea. The hot tea is initially at 80˚C. 1.4 kg of ice, initially at –10˚C, is placed in the tea and allowed to come to thermal equilibrium inside a thermos. You can assume the hot tea is essentially all water. What will be the final temperature of this system?

Answer 1

Answer:

$T_f=7.6619^\circ C$

Explanation:

Given:

• mass of the tea, $m_t=1.8kg$

• temperature of the tea, $T_t=80^\circ C$

• mass of ice, $m_i=1.4kg$

• temperature of the ice, $T_i=-10^\circ C$

We, know

• specific heat of tea(water), $c_t=4.184 J.kg^{-1}.K^{-1}$

• specific heat of ice, $c_i=2108 J.kg^{-1}.K^{-1}$

• Latent heat of fusion of ice, $L_i=336000 J.kg^{-1}$

Heat required for warming of ice upto 0°C from -10°C

$Q_i=m_i\times c_i\times \Delta T$

$Q_i=1.4\times 2108\times 10$

$Q_i=29512 J$

Heat released during the cooling of tea upto 0°C from 80°C

$Q_t=m_t\times c_t\times \Delta T$

$Q_t=1.8\times 4184\times 80$

$Q_t= 602496 J$

∵$Q_t >Q_i$

∴The heat from the quantity that remains after bringing the ice to 0°C will be used for melting the ice.

Heat remaining afterthis process:

$Q_0=(Q_t -Q_i)$= 572984 J

Amount of heat energy required by the total mass of ice to melt at 0°C:

$Q_L=m_i.L_i$

$Q_L=1.4\times 336000$

$Q_L=470400 J$

Now, the remaining of the heat will be used to raise the temperature of the tea (water) from 0°C to a higher temperature.

Heat remaining after melting the total ice into water at 0°C which will increases the temperature of water:

$Q_w= Q_0-Q_L$

$Q_w=572984-470400$

$Q_w=102584 J$

Now, the final temperature of the tea(water)rising from 0°C:

∵whole ice(along with tea) is now water with initial temperature $T_i$= 0°C

$Q_w=(m_i+m_t)\times c_t\times(T_f-T_i)$

where:

$T_f$= final temperature

$102584=(1.8+1.4)\times 4184\times (T_f-0)$

$T_f= \frac{102584}{3.2\times 4184}$

$T_f=7.6619^\circ C$