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At a local convenience store, you purchase a cup of coffee, but, at 98.4°C, it is too hot to drink. You add 48.7 g of ice that is −2.2°C to the 248 mL of coffee. What is the final temperature of the coffee?

Respuesta :

Answer:

[tex]T = 95.56 ^oC[/tex]

Explanation:

Initial temperature of coffee is given as

[tex]T_1 = 98.4^oC[/tex]

initial mass of the coffee is given as

[tex]V = 248 mL[/tex]

let say the density of coffee is same as that of water so we have

[tex]m = 248 g[/tex]

mass of ice added in it

[tex]m_1 = 48.7 g[/tex]

[tex]T_2 = -2.2 ^oC[/tex]

now here energy given by hot coffee = energy absorbed by ice

so we will have

[tex]m_1c_{ice}\Delta T_1 + m_1 L + m_1c_{water}\Delta T_2 = m c_{water}\Delta T_3[/tex]

here we have

[tex]48.7(2.1)(2.2) + 48.7(335) + 48.7(4.186)(T - 0) = 248(4.186)(98.4 - T)[/tex]

[tex]225 + 16314.5 + 203.86T = 102151.8 - 1038.1 T[/tex]

[tex]1242 T = 118691.3[/tex]

[tex]T = 95.56 ^oC[/tex]

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