Question

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?

Answer 1

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

`color(white)(ml)q_1 color(white)(mm)+color(white)(mll) q_2 color(white)(mm)= 0`

`m_1c_1ΔT_1 + m_2c_2ΔT_2 = 0`

In this problem, we have

`m_1color(white)(l) = ?`
`c_1color(white)(m) = "0.502 J·°C"^"-1""g"^"-1"`
`ΔT_1 = T_"f" - T_"i" = "(76.2 - 100.0) °C" = "-23.8 °C"`

`m_2color(white)(l) = "50.0 g"`
`c_2color(white)(m)= "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 = "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"`