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A 32.1-g ice cube at 0 °C is added to 120 g of water in a 64.2-g aluminum cup. The cup and the water have an initial temperature of 23.5 °C. Calculate the equilibrium temperature of the cup and its contents. Do not enter unit.

Respuesta :

Answer:

[tex]T = 3.5^0 C[/tex]

Explanation:

Heat given by water + cup  = Heat absorbed by the ice

here we can say that let the final temperature of the system is "T"

so we will have heat absorbed by the ice given as

[tex]Q = mL + ms\Delta T[/tex]

[tex]Q = (32.1)(335) + (32.1)(4.186)(T - 0)[/tex]

[tex]Q_{in} = 10753.5 + 134.4 T[/tex]

now we will have heat given by cup + water as

[tex]Q = m_w s_w(23.5 - T) + m_c s_c(23.5 - T)[/tex]

[tex]Q = 120(4.186)(23.5 - T) + 64.2(0.900)(23.5 - T)[/tex]

[tex]Q_{out} = 560.1(23.5 - T)[/tex]

now we have

[tex]Q_{in} = Q_{out}[/tex]

[tex]10753.5 + 134.4T = 560.1(23.5 - T)[/tex]

[tex]10753.5 + 134.4T = 13162.4 - 560.1 T[/tex]

[tex]694.5T = 2409[/tex]

[tex]T = 3.5^0 C[/tex]

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