Question

An aluminium vessel of mass 0.5 kg contains 0.2 kg of water at 20°C. A block of iron of mass 0.2 kg at 100°C is gently put into the water. Find the equilibrium temperature of the mixture. Specific heat capacities of aluminium, iron and water are 910 J kg−1 K−1, 470 J kg−1 K−1 and 4200 J kg−1 K−1 respectively.

Answer 1

Final equilibrium temperature of the mixture is  25° C

Explanation:

Given mass of aluminium =0.5 kg

Mass of water=0.2 kg

Mass of iron=0.2 kg

Temperature of aluminium and water

=20°=293° K

Temperature of iron =100° C=373 k

Specific heat of AI = 910 J/kg−K

Heat gain =0.5 × 610(T−293)+0.2×4200×(T−293)

= (T−293)(0.5×910+0.2×4200)

Heat lost =0.2×470×(373−T)

We know heat gain = Heat lost

= (T−293)(0.5×910+0.2×4200)

= 0.2×470×(373−T)

= (T−293)(455+840) = 94(373−T)

(T−293) 1295/94 = (273−T)[1295/94 = - 14]

(T−293) × 14=373−T

14 T−293×14 = 373−T

T = 4475/15 = 298.33 K ~298 K

T =  298 K - 273 K = 25° C

Final equilibrium temperature of the mixture is  25° C

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Answer 2

Explanation:

Given data  

aluminium mass = 0.5 kg

water mass= 0.2 kg

iron mass= 0.2 kg

aluminium Specific heat = $910 \mathrm{J} \mathrm{kg}^{-1} \mathrm{K}^{-1}$

iron’s Specific heat = $470 \mathrm{J} \mathrm{kg}^{-1} \mathrm{K}^{-1}$

water Specific heat = $4200 \mathrm{J} \mathrm{kg}^{-1} \mathrm{K}^{-1}$

Let the mixture equilibrium temperature be T.

Aluminium and water temperature $=20^{\circ} \mathrm{C}=273+20=.293 \mathrm{K}$

Iron Temperature$=100^{\circ} \mathrm{C}=273+100=373 \mathrm{K}$

Iron Heat lost, $\mathrm{H}_{1}=0.2 \times 470 \times(373-\mathrm{T})$

water Heat gained$=0.2 \times 4200 \times(T-293)$

iron Heat gained $=0.5 \times 910 \times(T-293)$

Maximum heat produced with water and iron,

$\mathrm{H}_{2}=0.5 \times 910(\mathrm{T}-293)+0.2 \times 4200 \times(\mathrm{T}-293)$

$\mathrm{H}_{2}=(\mathrm{T}-293)[0.5 \times 910+0.2 \times 4200]$

Heat gain equal to  Heat lost

$(\mathrm{T}-293)[0.5 \times 910+0.2 \times 4200]=0.2 \times 470 \times(373-\mathrm{T})$

$\begin{aligned}(\mathrm{T}-293)(455+840) &=94(373-\mathrm{T}) \\(T-293) \frac{1295}{94} &=373-T\end{aligned}$

$\begin{aligned}&14 T-293 \times 14=373-T\\&15 T=373+4102=4475\end{aligned}$

$T=\frac{4475}{15}=298.33 K \approx 298 K \mathrm{T}=(298-273)^{\circ} \mathrm{C}=25^{\circ} \mathrm{C} Finally temperature =25^{\circ} \mathrm{C}$