Question 9a717

Oct 10, 2016

The final temperature will be 57.6 °C.

Explanation:

Two heat transfers are involved.

$\text{heat lost by Al + heat gained by water = 0}$

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

m_1c_1ΔT_1 + m_2c_2ΔT_2 = 0

${m}_{1} = \text{3.90 g}$
${c}_{1} = \text{0.900 J°C"^"-1""g"^"-1}$
ΔT_1 = T_f - "99.3 °C"

m_2 = 1.0 color(red)(cancel(color(black)("cm"^3))) × "1.00 g"/(1.0 color(red)(cancel(color(black)("cm"^3)))) = "1.0 g"
${c}_{2} = \text{4.184 J°C"^"-1""g"^"-1}$
ΔT_2 = T_f - "22.6 °C"

q_1 = m_1c_1ΔT_1 = 3.90 color(red)(cancel(color(black)("g"))) × "0.900 J·°C"^"-1"color(red)(cancel(color(black)("g"^"-1"))) × (T_f -"99.3 °C")
$= 3.51 {T}_{f} \textcolor{w h i t e}{l} \text{J°C"^"-1" - "348.5 J}$

q_2 = m_2c_2ΔT_2 = 1.0 color(red)(cancel(color(black)("g"))) × "4.184 J·°C"^"-1"color(red)(cancel(color(black)("g"^"-1"))) × (T_f - "22.6 °C")
$= 4.184 \textcolor{w h i t e}{l} {T}_{f} \textcolor{w h i t e}{l} \text{J°C"^"-1" - "94.56 J}$

${q}_{1} + {q}_{2} = 3.51 {T}_{f} \textcolor{w h i t e}{l} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{J")))"°C"^"-1" -348.5 color(red)(cancel(color(black)("J"))) + 4.184 T_fcolor(white)(l)color(red)(cancel(color(black)("J")))"°C"^"-1" - 94.56 color(red)(cancel(color(black)("J}}}} = 0$

$443.1 = 7.694 \textcolor{w h i t e}{l} {T}_{f} \text{°C"^"-1}$

T_f = 443.1/("7.694 °C"^"-1") = "57.6 °C"

Check:

q_1 = m_1c_1ΔT_1 = 3.90 color(red)(cancel(color(black)("g"))) × "0.900 J"·color(red)(cancel(color(black)("°C"^"-1""g"^"-1"))) × ("57.6 -99.3") color(red)(cancel(color(black)("°C"))) = 3.51 × ("-41.7")"J" = "-146 J"

q_2 = m_2c_2ΔT_2 = 1.0 color(red)(cancel(color(black)("g"))) × "4.184 J"·color(red)(cancel(color(black)("°C"^ "-1""g"^"-1"))) × "(57.6- 22.6)" color(red)(cancel(color(black)("°C"))) = "241 J" - "94.56 J" = "146 J"#

${q}_{1} + {q}_{2} = \text{-146 J + 146 J} = 0$