# Question 3a1a1

Dec 27, 2016

The temperature of the prepared coffee is 99.5 °C.

#### Explanation:

There are three heat transfers involved in this problem.

$\text{Heat gained by coffee + heat gained by sugar + heat lost by water} = 0$

${q}_{1} + {q}_{2} + {q}_{3} = 0$

The formula for the heat gained or lost by a substance is

color(blue)(bar(ul(|color(white)(a/a)q = mcΔTcolor(white)(a/a)|)))" "

m_1c_1ΔT_1 + m_2c_2ΔT_2 + m_3C_3ΔT_3 = 0

In this problem,

${m}_{1} = \text{5.0 g}$; $\textcolor{w h i t e}{m l} {c}_{1} = \text{0.20 J·°C"^"-1""g"^"-1}$; color(white)(ll)ΔT_1 = T_"f" - T_"i" = T_"f"color(white)(l) "- 25.0 °C"

${m}_{2} = \text{15.0 g}$; $\textcolor{w h i t e}{l l} {c}_{2} = \text{0.30 J·°C"^"-1""g"^"-1}$; color(white)(ll)ΔT_2 = T_"f" - T_"i" = T_"f"color(white)(l) "- 25.0 °C"

${m}_{3} = \text{200.0 g}$; ${c}_{3} = \text{4.184 J·°C"^"-1""g"^"-1}$; ΔT_3 = T_"f" - T_"i" = T_"f"color(white)(l) "- 100.0 °C"

q_1 = m_1c_1ΔT_1 = 5.0 color(red)(cancel(color(black)("g"))) × "0.20 J·°C"^"-1"color(red)(cancel(color(black)("g"^"-1"))) × (T_"f"color(white)(l) "- 25.0 °C") = 1.0T_"f"color(white)(l) "J·°C"^"-1" "- 25.0 J"

q_2 = m_2c_2ΔT_2 = 15.0 color(red)(cancel(color(black)("g"))) × "0.30 J·°C"^"-1"color(red)(cancel(color(black)("g"^"-1"))) × (T_"f"color(white)(l) "- 25.0 °C") = 4.5T_"f" color(white)(l)"J·°C"^"-1" "- 112.5 J"

q_3 = m_3c_3ΔT_3 = 200.0 color(red)(cancel(color(black)("g"))) × "4.184 J·°C"^"-1"color(red)(cancel(color(black)("g"^"-1"))) × (T_"f"color(white)(l) "- 100.0 °C") = 836.8T_"f" color(white)(l)"J·°C"^"-1" "- 83 680 J"

${q}_{1} + {q}_{2} + {q}_{3} = 1.0 {T}_{\text{f"color(white)(l) "J·°C"^"-1" "- 25 J" + 4.5T_"f" color(white)(l)"J·°C"^"-1" "- 112.5 J" + 836.8T_"f" color(white)(l)"J·°C"^"-1" "- 83 680 J}} = 0$

842.3T_"f"color(red)(cancel(color(black)("J")))·"°C"^"-1"color(white)(l) "- 83 818" color(red)(cancel(color(black)("J"))) = 0#

${T}_{\text{f" = "83 818"/("842.3 °C"^"-1") = "99.5 °C}}$