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A sample of an unknown substance has a mass of 0.158 kg. If 2,510.0 J of heat is required to heat the substance from 32.0°C to 61.0°C, what is the specific heat of the substance?

Respuesta :

Given:

Mass of the unknown substance, m = 0.158 kg = 158 g

Heat absorbed Q = 2510.0 J

Initial temperature T1 = 32 C

Final temperature T2 = 61 C

To determine:

The specific heat of the substance

Explanation:

The heat absorbed to raise the temperature from T1 to T2 is-

Q = mcΔT

where c = specific heat

c = Q/mΔT = Q/m(T2-T1)

c = 2510 J/158 g * (61-32) C = 0.548 J/gC

Ans: Specific of the substance is 0.548 J/gC

Cricetus

The specific heat will be "0.548 J/gC".

According to the question,

  • Mass, [tex]m = 0.158 \ kg[/tex] or, [tex]158 \ g[/tex]
  • Heat absorbed, [tex]Q = 2510.0 \ J[/tex]
  • Initial temperature, [tex]T_1 = 32^{\circ} C[/tex]
  • Final temperature, [tex]T_2 = 61^{\circ} C[/tex]
  • Change in temperature, [tex]\Delta T = T_2 -T_1[/tex]

                                                         [tex]= 61-32[/tex]

                                                         [tex]= 29^{\circ} C[/tex]

As we know,

→ [tex]Q = mc \Delta T[/tex]

or,

→  [tex]c = \frac{Q}{m \Delta T}[/tex]

By putting the values,

      [tex]= \frac{2510}{158\times 29}[/tex]

      [tex]= \frac{2510}{4582}[/tex]

      [tex]= 0.548 \ J/gC[/tex]

Thus the response above is correct.

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