Question

If 200 ml of tea at 90 °C is poured into a 400 g glass cup initially at 25 °C, what would be the final temperature of the mixture?
Note: The specific heat capacity of water is 4186 J/kg °C, and the specific heat capacity of glass is 840 J/kg °C.

Answer 1

The final temperature of the mixture is 43.62 °C

To solve the question above, we apply the law of calorimetry

The law of calorimetry: which states that if there is no lost of heat the surrounding, heat lost is equal to heat gained, or it can be stated as heat absorbed by a cold body is equal to heat released by a hot body, provided there is no lost of heat to the surrounding.

The law above is expressed mathematically as

Cm(t₁-t₃) = C'm'(t₃-t₂)............. Equation 1

Using equation 1 to solve the question,

Let: C = specific heat capacity of glass cup, m = mass of glass cup, C' = specific heat capacity of tea, m' = mass of tea, t₁ = initial temperature of tea, t₂ = initial temperature of glass cup, t₃ = final temperature of the mixture.

From the question,

Given: m = 400 g = 0.4 kg, C = 840 J/kg°C, m' = 200g (tea is a liquid made of water and the volume of water in ml is thesame a its mass in gram) = 0.2 kg, C' = 4186 J/kg.°C, t₁ = 90°C, t₂ = 25°C

Substitute these values into equation 1 and solve for t₃

0.4(840)(90-t₃) = 0.2(4186)(t₃-25)

336(90-t₃) = 837.2(t₃-25)

30240-336t₃ = 837.2t₃-20930

collect like terms

837.2t₃+336t₃ = 30240+20930

  1173.2t₃ = 51170

t₃ = 51170/1173.2

t₃ = 43.62 °C

Hence, the final temperature of the mixture is 43.62 °C

Answer 2

Answer:

The final temperature of the system is 41.657 °C.

Explanation:

Let consider the tea-cup system as an isolated system, that is, a system with no energy and mass interactions with the surroundings. From the perspective of the First Law of Thermodynamics, the tea releases heat, which is received by the glass cup until thermal equilibrium is reached. The following formula represents the model under assumption that process was at steady state:

$\rho_{w}\cdot V\cdot c_{w}\cdot (T_{w,o}-T) + m_{g}\cdot c_{g}\cdot (T_{g,o}-T) = 0$ (1)

Where:

$\rho_{w}$ - Density of water, in grams per mililiter.

$V$ - Volume of tea, in mililiters.

$c_{w}$ - Specific heat of water, in joules per gram-degree Celsius.

$m_{g}$ - Mass of the glass cup, in kilograms.

$c_{g}$ - Specific heat of the glass cup, in joules per gram-degree Celsius.

$T_{w,o}$ - Initial temperature of tea, in degrees Celsius.

$T_{g,o}$ - Initial temperature of the glass cup, in degrees Celsius.

$T$ - Final temperature of the tea-cup system, in degrees Celsius.

If we know that $\rho_{w} = 1\,\frac{g}{mL}$, $V = 200\,mL$, $c_{w} = 4.186\,\frac{J}{g\cdot ^{\circ}C}$, $m_{g} = 400\,g$, $c_{g} = 0.840\,\frac{J}{g\cdot ^{\circ}C}$, $T_{w,o} = 90\,^{\circ}C$ and $T_{g,o} = 25\,^{\circ}C$, then the final temperature of the tea-cup system is:

$\rho_{w}\cdot V\cdot c_{w}\cdot (T_{w,o}-T) + m_{g}\cdot c_{g}\cdot (T_{g,o}-T) = 0$

$\rho_{w}\cdot V\cdot c_{w}\cdot T_{w,o} +m_{g}\cdot c_{g}\cdot T_{g,o} - (\rho_{w}\cdot V\cdot c_{w}+m_{g}\cdot c_{g})\cdot T = 0$

$T = \frac{\rho_{w}\cdot c_{w}\cdot T_{w,o}+m_{g}\cdot c_{g}\cdot T_{g,o}}{\rho_{w}\cdot V\cdot c_{w}+m_{g}\cdot c_{g}}$

$T = \frac{\left(1\,\frac{g}{mL} \right)\cdot \left(4.186\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (200\,mL)\cdot (90\,^{\circ}C)+(400\,g)\cdot \left(0.840\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (25\,^{\circ}C)}{\left(1\,\frac{g}{mL} \right)\cdot \left(4.186\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (200\,mL) +(400\,g)\cdot \left(0.840\,\frac{J}{g\cdot ^{\circ}C} \right)}$

$T = 41.657\,^{\circ}C$

The final temperature of the system is 41.657 °C.