A 40-g block of ice is cooled to -78 degrees C and is then added to 560 g of water in an 80-g copper calorimeter at a temperature of 25 degrees C. Determine the final temperature of the system consisting of the ice, water, and calorimeter. (If not all the ice melts, determine how much ice is left.) Remember that the ice must first warm to 0 degrees C, melt, and then continue warming as water. (The specific heat of copper and ice are 387 and 2090 J/kg degrees C, respectively. The latent heat of melting of ice is 3.33 x 105 J/kg.)

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rejkjavik rejkjavik · Answer 1 · 2019-10-17T12:23:41+02:00

Answer:

[tex]\theta = 15.56\ degree[/tex]

Explanation:

Heat required to heat ice is ΔQ1

ΔQ1 = msΔQ

[tex]= \frac{40}{1000} \times 2090 \times 78 = 6520.85 J[/tex]

Heat required to melt ice is ΔQ2

ΔQ2 = mL

[tex]= 40\times 334 = 13360 J[/tex]

Heat required to coll to o degree c is ΔQ3

ΔQ3 [tex]= m_w S_w \Delta Q + m_s S_s \Delta Q[/tex]

[tex]= 560\times 4.2\times 25 + 80\times 0.385 \times 25[/tex]

ΔQ3 = 59570 J

since ΔQ3 > ΔQ1 + ΔQ2

New temperature \theta at which ice melt to water

[tex]m_i S_i \Delta Q + m_i L + m_i s_w \theta = m_w S_w\Delta \theta + m_c S_c \Delta \theta[/tex]

[tex]\frac{40}{100}\times 2090\times 78 + 40\times 334 + 40\times 4.2 \theta = 560\times 4.2( 25-\theta) + 80\times 0.385(25-\theta)[/tex]

[tex]6520.8 + 13360 + 1680 = (2352 +30.8)(25 -\theta)[/tex]

solving for [tex]\theta[/tex] we get

[tex]\theta = 15.56\ degree[/tex]