Asample 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? use mc029-1.jpg.

The specific  heat of the substance  is  0.548  J/g/c

calculation

 Specific heat is calculated  using  Q=MCΔT  formula, where;

Q(heat)=  2510.0 j

 M (mass)=  0.158  kg  convert into  g

   = 1 Kg = 1000 g

     0.158 kg= ? g

      = (0.158 kg  x  1000g)  / 1  kg = 158  g

C(  specific heat capacity) = ?

ΔT(change in temperature)  = 61.0 -32c   = 29 c

   make  C  the subject of the  formula  by dividing both side by m andΔT

  c=  Q /mΔT

 C=  [2510.0 j  /    158 g x 29]  = 0.548  j/g/c

120,630,890

step-by-step explanation:

110+10=120,620+10=630,880+10=890

po-215 i got it right on the quiz