# Question bcfb3

Jan 15, 2017

Here's how you do it.

#### Explanation:

The guiding principle is the Law of Conservation of Energy: the sum of all the energy changes must add up to zero.

EXAMPLE

What is the final temperature if 10.0 kg of water at 10.0 °C is mixed with 2.00 kg of water at 100 °C? The specific heat capacity of water is $\text{4.184 kJ·°C"^"-1""kg"^"-1}$

Solution

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

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

where

$m$ is the mass of the substance.
$c$ is its specific heat capacity.
ΔT = T_"f" - T_"i" is the change in temperature.

In this problem, there are two heat flows.

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

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

m_1c_1ΔT_1 + m_2c_2ΔT_2 = 0

In this problem,

${m}_{1} = \text{10.0 kg";color(white)(mmm) m_2 = "2.00 kg}$

$\textcolor{w h i t e}{m m} {c}_{1} = {c}_{2} = \text{4.184 kJ·K"^"-1""kg"^"-1}$

ΔT_1 = T_"f"color(white)(l) "- 10.0 °C"; ΔT_2 = T_"f" color(white)(l)"- 100.0 °C"

m_1color(red)(cancel(color(black)(c_1)))ΔT_1 + m_2color(red)(cancel(color(black)(c_2)))ΔT_2 = 0

Since ${c}_{1} = {c}_{2}$, we can cancel them from the equation. This gives

m_1ΔT_1 + m_2ΔT_2 = 0

10.0 color(red)(cancel(color(black)("kg")))(T_"f"color(white)(l) "- 10.0 °C") + 2.00 color(red)(cancel(color(black)("kg")))(T_"f" color(white)(l)"- 100.0 °C"#

$10.0 {T}_{\text{f" - "100 °C" + 2.00T_"f" - "200 °C}} = 0$

$12.0 {T}_{\text{f" = "300 °C}}$

${T}_{\text{f" = "300 °C"/12.0 = "25.0 °C}}$