Using the Heat Equation

Question

What mass, in grams, of aluminum fins could 1673 J of energy heat from 13.74 ∘C to 21.17 ∘C ? Aluminum has a specific heat of 0.897 (J/g)⋅∘C

Express your answer to three significant figures.

Answer 1

Hey Marcella,

Our heat equation is Q=mcΔT where,

Q is the heat transfer in Joules.

m is the mass of the heating element in kg.

c is the specific heat capacity of the element.

ΔT is the change in temperature in C°

Looking at our question we are given the heat transfer. Q = 1673 J

we are also given the specific heat of Aluminum c = .897 (J/(g•C°)

Lastly we are given the temperature change ΔT = 21.17-13.74 = 7.43 C°

Looking at our heat equation Q=mcΔT, we are only missing m, the mass of the heating element.

Because we only have 1 unknown variable, we can solve for it by moving it to one side of the equation by itself.

In this case we divide both sides of the equation by c and ΔT to get the mass(m) by itself

The resulting equation is

m = Q ⁄ (c•ΔT)

plugging in our values for Q, c, and ΔT we find

m = 1673/(.897•7.43) = 251.02. Because the lowest sig fig value given is to 3 significant figures(specific heat of aluminum), our answer must be 3 sig figs as well.

Therefore our final answer is 251g of aluminum.

We can verify our approach by making sure our units line up in our heat equation. m = Q ⁄ (c•ΔT)

Plugging in the units for each variable and cancelling, we are left with mass = mass.

*Note that specific heats are usually reported in J/(kg•C°)