Specific Heat and Thermal Flow

Specific Heat

The heat required to raise the temperature of the unit mass of a given substance (usually one gram) by a given amount (usually one degree). The lighter the substance in terms of mass the faster it will cool

q= cm tThe heat added = (specific heat) (mass) (change in temperature)

q

c m t

The Reason the sand at the beach is hotter than the water

• Every substance has its own specific heat

Substance Specific Heat Jkg-1K-1

Aluminum 900

Brass 380

Copper 390

Platinum 130

Sand 800

Silver 230

Water 4100

The specific heat of sand is 800The specific heat of water is 4,100.Even though the sun is shining the same on the sand and on the water, it heats the sand up much quicker because the specific heat of the sand is lower Likewise, sand does not need to lose nearly as much energy as water to produce equivalent cooling. Since it "holds" a lot less energy, it cools down much faster this is true when you compare any two substances

What about other metals?

• If you are given a piece of each of the metals with the same mass and you heated them to 200* C

• And then dropped them into a beaker having 200ml 15*C water. Which metal would cool the fastest?

• IF you said Platinum you are right and it’s for the same reason as the sand it has a lower specific heat and will not “hold” the heat as long

Substance Specific Heat Jkg-1K-1

Aluminum 900

Brass 380

Copper 390

Platinum 130

Sand 800

Silver 230

Water 4100

• How much energy would be needed to raise the temperature of a 10kg block of copper by 20°C? • Looking at the specific heat of copper on the table above

= 390 Jkg-1K-1• Mass = 10kg • ΔT = 20°C

q = mc ΔT

q= 10(390)20 = 78000Joules or 78KJ

Example of a Problem for calculating Heat Energy

Example of a Problem for calculating Heat Energy

• 1. Gold has a specific heat of 0.129 J/(g×°C). How many joules of heat energy are required to raise the temperature of 15 grams of gold from 22 °C to 85 °C?

• you are solving for q = ? • You are given m = 15 g c = 0.129 J/g°C ΔT = 85°C - 22°C

= 63°C • To solve it you set up the equation • q = mc ΔT • q = (15 g)(0.129 J/g°C)(63 °C) • q = 121.9 J

Heat Flow

• Heat flow is a very simple concept with the exception that most people always think the process moves opposite. Heat ALWAYS transfers from warmer to cooler objects