Question

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

Answer 1

Answer:

(a) 0.2 J/g°K

(b) 24 J

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

Answer 2

Answer:

See below

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