yelitzamburgos06
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An unknown mass of aluminum requires 6290 joules to raise
the temperature from 23.9°C to 80.0°C. What is the mass of the
aluminum?
Specific heat value of aluminium is 0.900 J/gK

Respuesta :

Answer:

124.579 g

Explanation:

Amount of heat required to change the temperature of body is given by the equation given below

Q = m* c * ΔT ____________________equation A

where Q is the total heat energy required by the any object

c is the specific heat capacity of the body J/gK

ΔT is the difference of final temperature of the object and initial temperature of the object

______________________________________

Given Q = 6290 joules

c = 0.900 J/gK

To calculate the temperature in kelvin from Celsius

we can use the formula

K = C + 273

initial temperature  = 23.9°C

initial temperature on kelvin scale =  23.9° + 273 = 296.9

Final temperature  = 80°C

initial temperature on kelvin scale =  80° + 273 = 353

ΔT(temperature difference) = 353 - 296.9 = 56.1

___________________________________________

let the mass of of the  aluminum be m g

substituting the value of Q , c, ΔT in equation A we have

Q = m* c * ΔT

=> 6290 = m * 0.900 * 56.1

=> m = 6290/(0.900 * 56.1 )

=> m = 6290/50.49

=> m = 124.579 g

The mass of the  aluminum is 124.579 g

CintiaSalazar

Taking into account the definition of calorimetry, the mass of the aluminum is 124.58 grams.

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

Sensible heat is the heat energy that is supplied to a body or an object, causes its temperature to increase without affecting its molecular structure and therefore its phase. In other words, sensible heat is the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

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= 6290 J
  • c= 0.9 [tex]\frac{J}{gK}[/tex]
  • m=?
  • ΔT= Tfinal - Tinitial= (80 - 23.9) C= 56.1 C= 56.1 K  Being a temperature difference, the value is the same in ° C and in ° K

Replacing:

6290 J= 0.9[tex]\frac{J}{gK}[/tex]× m× 56.1 K

Solving:

6290 J= 50.49 [tex]\frac{J}{g}[/tex]× m

6290 J÷ 50.49 [tex]\frac{J}{g}[/tex]= m

124.58 g= m

Finally, the mass of the aluminum is 124.58 grams.

Learn more:

  • brainly.com/question/11586486?referrer=searchResults
  • brainly.com/question/24724338?referrer=searchResults

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