Calorimetry?

Question

The specific heat capactiy of silver is 0.24 J/g 'C.

a) Calculate the energy required to raise the temperature of 150.0 g Ag from 273 K to 298 K.

b) Calculate the energy required to raise the temperature of 1.0 mol Ag by 1.0'C (called the molar heat capacity of silver).

Answer 1

∆H = mCp∆T

1) Tfinal 25C, Tinitial - 0C ∆T = Tfinal - Tinitial = 25ºC

∆H = 150 g * 0.24 J/gºC * 25ºC = 900J

2) Ag MW = 107.8682(2)  g/mole

∆H/mole = 107.9 g * 0.24 J/gºC * 1ºC = 25.896 J

so the molar heat capacity is 25.89 J/moleºC

in wiki it's 25.35 J/moleºC

ref: http://en.wikipedia.org/wiki/Silver

Answer 2

a)
∆Q = mc∆T
∆Q = (150.0g )(0.24 J/gC°)(298K - 273K)
∆Q = 900 J

where 1C° = 1 K


b)
∆Q = mc∆T
n = m/M
m = nM

where
n = no. of moles
m = the given mass
M = molecular mass For silver it's 107.86 g/mole

∆Q = m c∆T
∆Q = nM c∆T
∆Q = (1.0 mole x107.86 g/mole) (0.24 J/g C°) (1.0 C°)
∆Q = 25.8864 J
∆Q ≈ 25.9 J

Sorry for the delay in looking for the molecular mass of silver.

Answer 3

a) Q = mC(T2-T1)
Q = (150 g)(0.24 J/g K)(298 -273K) = 900 J

b) 1 mol Ag = 107.86 g
Q = (107.86 g/mol)(0.24 J/g°C)(1°C) = 25.88 J / mol

Answer 4

a) Use the equation q= m c DT

b. Multiply the specific heat by the molar mass of silver.

Answer 5

a) 298K - 273K = 25degK difference = 25degC difference

150.0gAg x 25degC x 0.24J/g-degC = 900 J

b) Atomic weight: Ag=107

107gAg/1molAg x 1.0-degC x 0.24J/g-degC = 28.7J/mol to three significant figures