A68-g sample of sodium is at an initial temperature of 42 °c. if 1840. joules of heat are applied to the sample, what is the final temperature of the sodium?

Final T = 64.0°C.

Explanation:

Q = m.c.ΔT.

where, Q is the amount of heat absorbed by Na (Q = 1840 J),

m is the mass of Na (m = 68.0 g),

c is the specific heat capacity of Na (c = 1.23 J/g °C),

ΔT is the temperature difference (final T - initial T) (ΔT = final T - 42.0°C).

∵ Q = m.c.ΔT.

∴ (1840 J) = (68.0 g)(1.23 J/g °C)(final T - 42.0°C)

(final T - 42.0°C) = (1840 J)/(68.0 g)(1.23 J/g °C) = 22.0°C.

∴ final T = 22.0°C + 42.0°C = 64.0°C.

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