Question 91423

May 2, 2017

You need to use 100. g of ice.

Explanation:

There are two heat transfers involved.

$\text{heat gained by the ice + heat lost by the drink = 0}$

$\textcolor{w h i t e}{m m m m m m} {q}_{1} \textcolor{w h i t e}{m m m l l} + \textcolor{w h i t e}{m m m m m} {q}_{2} \textcolor{w h i t e}{m m m m l} = 0$

color(white)(mmmm)m_1Δ_text(fus)H color(white)(mml)+ color(white)(mmml)m_2CΔT color(white)(mmll)= 0

Let's calculate these heats separately.

m_1color(white)(mll) = ?
Δ_text(fus)H = "0.33 kJ/g"

q_1 = m_1 × "0.33 kJ/g" = 0.33m_1 color(white)(l)"kJ/g"

I assume that the drink has the same density as water. Then

m_2 = 12.0 color(red)(cancel(color(black)("oz"))) × (29.57 color(red)(cancel(color(black)("mL"))))/(1 color(red)(cancel(color(black)("oz")))) × "1.00 g"/(1 color(red)(cancel(color(black)("mL")))) = "354.8 g"

ΔT = T_2 - T_1 = "(35 - 75) °F" = "-40" color(red)(cancel(color(black)("°F"))) × "5 °C"/(9 color(red)(cancel(color(black)("°F")))) = "-22.2 °C"

q_2 = 354.8 color(red)(cancel(color(black)("g"))) × 4.18 color(white)(l)"J"·color(red)(cancel(color(black)("g"^"-1""°C"^"-1"))) × "-22.2" color(red)(cancel(color(black)("°C"))) = "-32 920 J" = "-32.92 kJ"

Now, we can add the two heats.

${q}_{1} + {q}_{2} = 0.33 {m}_{1} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{kJ")))·"g"^"-1" - 32.92 color(red)(cancel(color(black)("kJ}}}} = 0$

m_1 = 32.92/("0.33 g"^"-1") = "100. g"#