Will mark time is running ! what is the specific heat of a substance if 300 j are required to raise the temperature of a 267–g sample by 12ºc? (must show all work and include units of measure to receive full credit.)

We can use the heat equation,
Q = mcΔT 

Where Q is the amount of energy transferred (J), m is the mass of the substance (kg), c is the specific heat (J g⁻¹ °C⁻¹) and ΔT is the temperature difference (°C).

According to the given data,

Q = 300 J 

m = 267 g

c = ?
ΔT = 12 °C

By applying the formula,

300 J = 267 g x c x 12 °C
       c = 0.0936 J g⁻¹ °C⁻¹

Hence, specific heat of the given substance is 0.0936 J g⁻¹ °C⁻¹.

each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i