Question

A 328-g sample of water absorbs 5.78 × 103 J of heat. Calculate the change in temperature for the water. If the water is initially at 25.0 °C, what is its final temperature?

Answer 1

The change in temperature for the water is 4.21 C and its final temperature is 29.21 C

Step-by-step explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

There is a direct proportional relationship between heat and temperature. The constant of proportionality depends on the substance that constitutes the body as on its mass, and is the product of the specific heat by the mass of the body. So, the equation that allows calculating heat exchanges is:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case, you know:

• Q= 5.78*10³ J
• c= 4.186
  `(J)/(g*C)`
• m= 328 g
• ΔT= Tfinal-Tinitial

Replacing:

5.78*10³ J= 4.186
`(J)/(g*C)` *328 g* ΔT

Solving:

ΔT=
`(5.78*10^(3) J )/(4.186(J)/(g*C)*328 g)`

ΔT= 4.21 C

Being ΔT= Tfinal - Tinitial and Tinitial=25 C, then:

4.21 C= Tfinal - 25 C

4.21 C + 25 C= Tfinal

29.21 C= Tfinal

The change in temperature for the water is 4.21 C and its final temperature is 29.21 C