What is the mass of a sample that absorbs 32J of energy when it is heated from 274K to 314K and has a specific heat of 0.20 J/gK. *

The answer to your question is 4 g

Explanation:

Data

mass = m = ?

Energy = E = 32 J

Temperature 1 = T1 = 274°K

Temperature 2 = T2 = 314°K

Specific heat = C = 0.20 j/g°K

Formula

     E = mC(T2 - T1)

-Solve for m

     m = E / C(T2 - T1)

-Substitution

      m = 32 / 0.2(314 - 274)

-Simplification

     m = 32 / 0.2(40)

     m = 32 / 8

-Result

     m = 4 g

-Conclusion

   The sample has a mass of 4 g.

Mass of sample is 4 grams

Explanation:

specific heat = s= 0.20 J / gk

ΔT =T₂ - T₁

Where T₁ = 274 k

            T₂ = 314 k

  ΔT = 314 - 274

        =  40 K

Energy absorbed (q) = 32 J

q = msΔT

Make m the subject of formula

m = q / sΔT

   = 32 / 0.2 x 40

   = 32 / 8

   = 4 grams

Therefore, the mass of sample = 4 grams