Question

4.05 g of methane is burned and all of the heat is from this is absorbed by 1.0x10^3 g of 20.0 degree C water (which has a specific heat of 4.18Jg^-1 C^-1). What is the final temperature of the water?

Answer 1

CH4(g)+2O2(g)--->CO2(g)+2H2O(l) Delta H= -890.2 kJ

For 1 mole CO2 890.2 kJ heat was released

Hence for 4.05 gm = 890.2 x 4.05/16 = 225.33 kJ

Q = m c ∆T
Q = quantity of heat in joules (J) 225.33 x 10^3 J
m = mass of the substance acting as the environment in
grams (g) 1000g
c = specific heat capacity (4.19 for H2O) in J/(g ^oC)
∆T = change in temperature = Tfinal - Tinitial in ^oC

225.33 x 10^3 = 1000 x 4.184 x (Tfinal - 20)

Tfinal = 53.85 + 20 = 73.85 ^oC

final temperature of the water = 73.85  ^oC