Question

How many moles of butane must burn to increase the temperature of 10dm³ of water from 30⁰ C to 100^o C? Given that dH_(comb) of butane, density of water and specific heat of water are -2.89*10³ KJ/mol, 1.0 g/cm³ .4.184JK ^-1 g ^-1 .

Answer 1

Volume = 10 dm³

1 dm = 10cm

so, volume= 10⁴cm³

mass = volume x density

= 10⁴ x 1

= 10⁴g

c = 4.184 JK^-1g^-1

ΔT = T_f-T_i

= 373-303

= 70 K

q = mcΔT

= 10⁴ x 4.184 x 70 J

For butane ΔH = 2.89 x 10⁶ J/mole

2.89 x 10⁶ J is obtained by = 1 mole of butane

1 J is obtained by = 1/2.89 x 10⁶ mole

10⁴ x 4.184 x 70 J is obtained by= 1 x 10⁴ x 4.184 x 70 /2.89 x 10⁶ mole

= 101.34 x 10^-2 mole