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

`T_f=7.6619^\circ C`

Step-by-step 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* c_i* \Delta T`

`Q_i=1.4* 2108* 10`

`Q_i=29512 J`

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

`Q_t=m_t* c_t* \Delta T`

`Q_t=1.8* 4184* 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* 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)* c_t*(T_f-T_i)`

where:

`T_f`= final temperature

`102584=(1.8+1.4)* 4184* (T_f-0)`

`T_f= (102584)/(3.2* 4184)`

`T_f=7.6619^\circ C`