Question

how do i caculate the trmperature change that 725 grams of aluminum will undergo when 2.35x10^4 Joules of thermal energy are added to it?

Answer 1

36°C

Step-by-step explanation:

Given parameters:

Mass of aluminum = 725g

Quantity of heat = 2.35 x 10⁴J

Unknown:

Temperature change = ?

Solution:

To solve this problem, we simply use the expression below:

The quantity of energy is given as:

Q = m C Δt

Q is the quantity of energy

m is the mass

C is the specific heat capacity of aluminum = 0.9J/g°C

Δt is the change in temperature

The unknown is Δt;

Δt =
`(Q)/(mc)` =
`(2.35 x 10^(4) )/(725 x 0.9)` = 36°C