Question

A 30.5 g sample of an alloy at 93.0 oC is placed into 50.0 g of water (cH2O = 4.184 J g−1 oC−1) at 22.0 oC in an insulated coffee cup with a heat capacity of 9.2 J oC−1. If the final temperature is

31.1 oC, what is the specific heat capacity of the alloy?

1. A) 851 J g−1 oC−1
2. B) 0.367 J g−1 oC−1
3. C) 0.218 J g−1 oC−1
4. D) 0.553 J g−1 oC−1
5. E) 1.05 J g−1 oC−1

Methane (CH4) is a common fuel to heat homes in the winter. What is the molar enthalpy of

combustion of methane? Assume this combustion occurs entirely in the gas phase.

Bond Enthalpies (in kJ mol−1):

C−C: 347  C=C: 614

C−H: 413  C=O: 745

H−H: 432 O=O: 498

O−H: 467

a)−710 kJ mol−1 

b)−297 kJ mol−1

c)−1843 kJ mol−1 

d)+792 kJ mol−1 

e)+567 kJ mol−1

Answer 1

=>

Heat lost = Heat gained

$m_wc_w∆T + c_c∆T= m_ac_a∆T$

$50 × 4.184 × (31.1 - 22) + 9.2× (31.1-22)= 30.5 ×c_a× (93- 31.1)$

$1987.44=1887.95× c_a$

$c_a = \dfrac{1987.44}{1887.95} =1.053\ J/kg°C$

=>

$CH_4+ 2O_2 \to CO_2 + 2H_2O$

. Methane has only 4 C-H bonds

= 4(413) = 1652 KJmol^-1

. Oxygen has 1 O=O bond

= 2(498) = 996 KJmol^-1

. Carbon (iv) Oxide has 2 C=O bonds

= 2(745) = 1490 KJmol^-1

. Water has 2 O-H bonds

= 4(467) = 1868 KJmol^-1

∆H_product - ∆H_reactant = ∆H_combustion

$∆H_{combustion }= 1868 + 1490- (996+1652) = 710\ kJmol^{-1}$