Question

When one mole of C6H6 is burned, 3.27 MJ of heat is produced. When the heat from burning 7.19 g of C6H6 is added to 5.69 kg of water at 21.0°C, what is the final temperature of the water?

Answer 1

Given : Heat produced by burning one mole of C₆H₆ = 3.27 MJ , mass of C₆H₆= 7.19 g , mass of water (m)= 5.69 kg , initial temperature of water(T_i)= 21°c

Constant : molar mass of C₆H₆ (M)= 78.11 g/mol, specific heat of water (c_p)= 4.186 J/g°c

Solution:

Heat produced by burning of one mole of C₆H₆ = 3.27 MJ

Now heat produced by burning of 7.19 g of C₆H₆ is given by=

Q = [7.19g ÷78.11 g/mol]*3.27 ×10⁶ J = 301002.43 J

Now heat transfered to the water is given by:

Q = mc_pT

301002.43 J = (5690 g)(4.186 J/g°c)(T_f - 21°c)

T_f - 21 = 12.64

T_f = 12.64+21= 33.64°c

Answer : final temperature of water (T_f)= 33.64°c