abbie3935
contestada

What is the specific heat of an unknown substance if a 2.50 g sample releases 12 calories as its
temperature changes from 25°C to 20°C?

Respuesta :

Neetoo

Answer:

c = 4016.64 j/g.°C

Explanation:

Given data:

Mass of substance = 2.50 g

Calories release = 12 cal (12 ×4184 = 50208 j)

Initial temperature = 25°C

Final temperature = 20°C

Specific heat of substance = ?

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Solution:

Q = m.c. ΔT

ΔT  = T2 - T1

ΔT  = 20°C - 25°C

ΔT  = -5°C

50208 j = 2.50 g . c. -5°C

50208 j = -12.5 g.°C .c

50208 j /-12.5 g.°C =  c

c = 4016.64 j/g.°C

CintiaSalazar

Taking into account the definition of calorimetry, the specific heat of the unknown substance is 0.96 [tex]\frac{calories}{gC}[/tex].

Calorimetry

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

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where

  • Q is the heat exchanged by a body of mass m.
  • c: specific heat substance.
  • ΔT is the temperature variation.

Specific heat of an unknown substance

In this case, you know:

  • Q= - 12 calories
  • c= ?
  • m= 2.50 g
  • ΔT= Tfinal - Tinitial= 20 C - 25 C= - 5 C

Replacing in the expression to calculate heat exchanges:

-12 calories= c× 2.50 g× (-5 C)

Solving:

-12 calories÷ [2.50 g× (-5 C)]= c

0.96 [tex]\frac{calories}{gC}[/tex]= c

Finally, the specific heat of the unknown substance is 0.96 [tex]\frac{calories}{gC}[/tex].

Learn more about calorimetry:

brainly.com/question/11586486?referrer=searchResults

brainly.com/question/24724338?referrer=searchResults