The specific heat of a certain type of metal is 0.128 J/(g⋅∘C). What is the final temperature if 305 J of heat is added to 79.4 g of this metal, initially at 20.0 ∘C?

Question

please help!! I am having a lot of trouble with this question

Answer 1

Hi, Isabella,

The specific heat unit is a favorite of mine. It took some convincing at first, but if you look at the units, a lot becomes clearer. Units are J/g^oC in this case. [Be aware that specific heat can be expressed in MANY different types of units. Moles instead of grams, calories in place of Joules, F instead of C, etc.]

But in all cases, it tells us how much heat is needed (J) to raise a specific amount of material (1 gram), by 1 degree (C, F, or K). If the specific heat of my pencil is 5J/g^oC, I know that I'd have to apply 5 Joules of energy for every gram of pencil, to raise the whole pencil 1 degree. If I had 2 grams, I'd need 10 Joules. If you keep the units in the calculations, it helps make everything clearer.

The specific heat of the metal in the problem is 0.128 J/g^oC. That's a relatively low number. Compare that to the specific heat of water, which is 4.186 J/g^oC. It takes (4.186/0.128) times as much energy to raise 1 gram of water as 1 gram of this metal. That is typical of metals. That's why we use them for cooking: the heat doesn't want to linger - the pan gets very hot quickly and the food is what gets most of the energy, especially the water.

The basic format to answer this question is the equation c = q/(m(T1 - T2)), where q is the heat, c is the specific heat, m is the mass, and (T1 - T2) is the temperature change (T1 is the initial temp). In this problem we know q, c and m. We want the temperature change. So let's rearrange the equation, first:

(T2 - T1) = q/mc

Plug in the values:

(T2 - T1) = 305J/(0.128J/g^oC * 79.4g)

Joules and grams cancel and ^oC moves to the top. An encouraging sign: We get the unit we want, ^oC.

(T2 - 20) = 30.01^oC

T2 = 50.01^oC , 50.0 ^oC with 3 sig figs

I hope this helps. Always keep units in the equations and cancel them methodically. That will help insure you've approached the problem correctly. Many problems sneak in different units to see if 1) you spot the deception, and 2) if you can adjust the calculations to match the units. For example, water's specific heat in various units in addition to the one above include: 76 J/moleK, 76 kJ/kgK, 1.0073 kcal/kgK. These are all equivalent to each other - only the units are different.

Beware, then enjoy.

Bob