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A 32.5 g cube of aluminum initially at 45.8 °C is submerged into 105.3 g of water at 15.4 °C. What is the final temperature of both substances at thermal equilibrium? (Assume that the aluminum and the water are thermally isolated from everything else.)

Respuesta :

Answer:

[tex]T = 17.26 ^oC[/tex]

Explanation:

At thermal equilibrium we have heat given by aluminium must be equal to the heat absorbed by the water

so we will have

[tex]Q_1 = Q_2[/tex]

[tex]m_1s_1\Delta T_1 = m_2s_2\Delta T_2[/tex]

so we will have

[tex]32.5(900)(45.8 - T) = 105.3(4186)(T - 15.4)[/tex]

so we have

[tex](45.8 - T) = 15.1(T - 15.4)[/tex]

so we have

[tex]16.1 T = 277.87[/tex]

[tex]T = 17.26 ^oC[/tex]

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