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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?

Sagot :

quasarJose

Answer:

36°C

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  = [tex]\frac{Q}{mc}[/tex]    = [tex]\frac{2.35 x 10^{4} }{725 x 0.9}[/tex]    = 36°C

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