Respuesta :
Answer:
La temperatura final es de aproximadamente 159,94°C
Explanation:
Los parámetros dados son;
La masa de la llave española de acero, m₁ = 200 gramos
La temperatura de la llave, T₁ = 550 ° C
La masa del recipiente de aluminio que contiene agua, m₂ = 250 gramos
La masa del agua en el recipiente de aluminio, m₃ = 220 gramos
La capacidad calorífica específica del hierro, [tex]C_{planchar}[/tex], c₁ = 0.499 ca/(g·°C)
La capacidad calorífica específica del aluminio, [tex]C_{Aluminio}[/tex], c₂ = 0.217 cal/(g·°C)
La capacidad calorífica específica del agua, [tex]C_{Agua}[/tex], c₃= 1 cal/(g·°C)
En equilibrio térmico, tenemos;
m₁·c₁·(T₁ - T) = m₂·c₂·(T -T₂) + m₃·c₃·(T - T₂)
Conectando los valores, da;
200 × 0.499 × (550 - T) = 250 × 0.217 × (T -18) + 220 × 1 × (T - 18)
Simplificando, usando una calculadora gráfica, obtenemos;
[tex]\dfrac{274450-499\cdot T}{5} = \dfrac{1097 \cdot T-19746}{4}[/tex]
De también encontramos 'T' al convertirlo en el tema de la ecuación anterior aún usando una calculadora gráfica;
T = 1196530/7481 °C ≈ 159.94°C
La temperatura final,T ≈ 159.94°C.
The final temperature, when thermal equilibrium is reached is 144°C.
Given the following data:
- Mass of key = 200 grams
- Final temperature of key = 550°C
- Mass of aluminum = 250 grams
- Mass of water = 220 grams
- Initial temperature of water = 18°C
- Specific heat capacity of key = 0.499 cal/g°C
- Specific heat capacity of aluminum = 0.217 cal/g°C
- Specific heat capacity of water = 1 cal/g°C
To determine the final temperature, when thermal equilibrium is reached:
Mathematically, heat capacity or quantity of heat is given by the formula;
[tex]Q = mc\theta[/tex]
Where:
- Q represents the quantity of heat.
- m represents the mass of an object.
- c represents the specific heat capacity.
- ∅ represents the change in temperature.
At an equilibrium state, the quantity of heat for the three substances is given by the equation:
[tex]M_kC_k(\theta_2 - \theta_1) = M_aC_a(\theta_2 - \theta_1) + M_wC_w(\theta_2 - \theta_1)\\\\200 \times 0.499 \times (500 - \theta_2) = 250 \times 0.217 \times (\theta_2 - 18) + 220 \times 1 \times (\theta_2 - 18)\\\\99.8(500 - \theta_2) = 54.25(\theta_2 - 18) + 220(\theta_2 - 18)\\\\49900 -99.8\theta_2 = 54.25\theta_2 - 976.5 + 220\theta_2 - 3960\\\\220\theta_2 + 99.8\theta_2 + 54.25\theta_2 = 49900 + 3960\\\\374.05\theta_2 = 53860\\\\\theta_2 = \frac{53860}{374.05}[/tex]
Final temperature, [tex]\theta_2[/tex] = 143.99 ≈ 144°C
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