# A sample of metal with a specific heat of .502 J/g°c is heated to 100.0°C and then placed a 50.0 g sample of water at 20.0°C. The final temperature of the system is 76.2°C. What is the mass of the metal used in process?

Feb 9, 2017

The mass of the metal is 459 g.

#### Explanation:

We must identify the heat transfers that are happening here.

One is the heat transferred from the metal as it cools (${q}_{1}$).

The second is the heat transferred to the water as it warms (${q}_{2}$).

Per the Law of Conservation of Energy, the sum of the two heat transfers must be zero.

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

The formula for the heat absorbed by or released from a substance is

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

where

$q$ is the quantity of heat
$m$ is the mass of the substance
$c$ is the specific heat capacity of the material
ΔT is the temperature change

This gives us

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

m_1c_1ΔT_1 + m_2c_2ΔT_2 = 0

In this problem, we have

m_1color(white)(l) = ?
${c}_{1} \textcolor{w h i t e}{m} = \text{0.502 J·°C"^"-1""g"^"-1}$
ΔT_1 = T_"f" - T_"i" = "(76.2 - 100.0) °C" = "-23.8 °C"

${m}_{2} \textcolor{w h i t e}{l} = \text{50.0 g}$
${c}_{2} \textcolor{w h i t e}{m} = \text{4.184 J·°C"^"-1""g"^"-1}$
ΔT_2 = T_"f" - T_"i" = "(76.2 - 20.0) °C" = "26.2 °C"

q_1 = m_1 × "0.502 J"·color(red)(cancel(color(black)("°C"^"-1")))"g"^"-1" × ("-23.8" color(red)(cancel(color(black)("°C")))) = "-11.95"color(white)(l)m_1 color(white)(l)"J·g"^"-1"

${q}_{2} = \text{50.0 g" × "4.184 J"·color(red)(cancel(color(black)("°C"^"-1")))"g"^"-1" × 26.2 color(red)(cancel(color(black)("°C"))) = "5481 J}$

q_1 + q_2 = "-11.95"m_1 color(red)(cancel(color(black)("J")))·"g"^"-1" + 5481 color(red)(cancel(color(black)("J"))) = 0

m_1 = 5481/("11.95 g"^"-1") = "459 g"