Calculate the total quantity of heat required to convert 25.0 g of liquid ccl₄(l) from 25.0°c to gaseous ccl₄ at 76.8°c (the normal boiling point for ccl₄)? the specific heat of ccl₄(l) is its heat of fusion is and its heat of vaporization is
a. 0.896 kj
b. 1.43 kj
c. 5.74 kj
d. 6.28 kj.

d. 6.28 kJ.

Explanation:

step by step explanation

Q=Q1+Q2

Q=mc(t1-t0) + mlv

Q=total heat generated

Q1=quantity of heat required to raise 25.0°C to 76.8°C to liquid state

Q2= quantity of heat required to convert ccl at 76.8°C to its gaseous state

M=mass of ccl4

C=specific heat capacity of ccl4(0.866j/gk)

T1=final temperature 76.8°C

T0=initial temperature  25.0 °C

LV=latent heat of vapourization (1.959 *10^2j/g)

Q=25*0.866 *(76.8-25) +25*1.959*10^2

Q=6.28kj

I think it might be c.

Explanation:

Hope it's right :3