# A 77 g sample of water with a temperature of 30°C is added to 370 g water at 92°C in an insulated container. What is the final temperature after thermal equilibrium is reached?

Nov 6, 2016

The final temperature is 81.3 °C.

#### Explanation:

There are two heat transfers involved in this problem.

$\text{Heat gained by cold water + heat lost by hot water} = 0$

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

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_1color(red)(cancel(color(black)(c)))ΔT_1 + m_2color(red)(cancel(color(black)(c)))ΔT_2 = 0

Since $c$ is a constant, we can cancel it and the equation becomes

m_1ΔT_1 = m_2ΔT_2

In this problem,

${m}_{1} = \text{77 g}$
ΔT_1 = T_"f" - T_1^° = T_"f"color(white)(l) "- 30 °C"
${m}_{2} = \text{370 g}$
ΔT_2 = T_"f" - T_2^° = T_"f"color(white)(l)_1color(red)(cancel(color(black)(c)))ΔT_1 + m_2color(red)(cancel(color(black)(c)))ΔT_2 = 0

77 color(red)(cancel(color(black)("g"))) × (T_"f"color(white)(l) "- 30 °C) + 370"color(red)(cancel(color(black)("g")))× (T_"f"color(white)(l)"- 92 °C") = 0

$77 {T}_{\text{f"color(white)(l)color(white)(l)"- 2310 °C" + 370T_"f"color(white)(l) "- 34 040 °C}} = 0$

$447 {T}_{\text{f" = "(2310 + 34 040) °C" = "36 350 °C}}$

${T}_{\text{f" = "36 350 °C"/447 = "81.3 °C}}$

The final temperature is 81.3 °C.