Question

When 28 grams of a metal are heated from 14.4°C to 150°C it takes 23,200). What is the specific heat of the metal?

Answer 1

6.11 J/g°C

Step-by-step explanation:

q = mc(delta theta)

23200 = 28 × c × (150 - 14.4)

23200 = 28 × c × 135.6

c = 23200/(28 × 135.6)

c = 6.110408765

Answer 2

6.11 J/g°C

Step-by-step explanation:

We need to use the heat equation: q = mCΔT, where q is the amount of heat required in Joules, m is the mass in grams, C is the heat capacity, and ΔT is the change in temperature.

Here, we know q = 23,200 Joules, m = 28 grams, and the ΔT is just the difference of the temperatures: 150 - 14.4 = 135.6°C. Substitute these values to find C:

q = mCΔT

23,200 = 28 * C * 135.6

C = 6.11

Thus, the specific heat is 6.11 J/g°C.