Question

1. A 4.0-g sample of glass was heated from 274K to 314 K. a temperature increase of 40 K, and was

found to have absorbed 32J of energy as heat.
a. What is the specific heat of this type of glass?
b. How much energy will the same glass sample gain when it is heated from 314 K to 344 K?

Answer 1

See below

Step-by-step explanation:

Q = m c T c = specific heat T = temp change Q = heat joules

32 = 4 c 40 c = .2 J/g-C

314 to 344 k is a change of 30 K

Q = m c T

= 4 * .2 * 30 = 24 Joules

Answer 2

(a) 0.2 J/g°K

(b) 24 J

Step-by-step explanation:

(a)

To find the specific heat capacity, you need to use the following equation:

Q = mcΔT

In this formula,

-----> Q = heat energy (J)

-----> m = mass (g)

-----> c = specific heat capacity (J/g°K)

-----> ΔT = change in temperature (K)

You can plug the given values into the equations and simplify to find the missing value.

Q = 32 J c = ? J/g°K

m = 4.0 g ΔT = 40 K

Q = mcΔT <----- Equation

32 J = (4.0 g) x c x (40 K) <----- Insert variables

32 J = (160) x c <----- Multiply 4.0 and 40

0.2 = c <----- Divide both sides by 160

(b)

To find the energy of the same sample, you can use the same equation. This time, you know the specific heat capacity, have a different change in temperature, and are solving for energy (Q).

Q = ? J c = 0.2 J/g°K

m = 4.0 g ΔT = 344 K - 314 K = 30 K

Q = mcΔT <----- Given equation

Q = (4.0 g)(0.2 J/g°K)(30 K) <----- Insert values

Q = 24 <----- Multiply