Question

The specific heat capacity of silver is 0.24 J/°C ·g.

Answer 1

Answer:

A) 900 J

B) 27.96 J

C) 1,628 J ≅ 1.63 kJ

Explanation:

The heat absorbed by the metal (silver) - or energy required to heat it -  is calculated as:

heat = mass x Cp x ΔT

Where Cp is the heat capacity (0.24 J/°C ·g) and ΔT is the change in temperature (final T - initial T).

A) Given:

mass = 150.0 g

final T = 298 K = 25°C

initial T = 273 K = 0°C

We calculate the energy in J to raise the temperature:

heat = mass x Cp x (final T - initial T)

       = 150 .0 g x 0.24 J/°C ·g x (25°C - 0°C )

       = 900 J

B) Given:

moles Ag= 1.0 mol

ΔT = 1.08°C

We first calculate the mass of silver (Ag) by multiplying the moles of Ag by the molar mass of Ag (MM = 107.9 g/mol)

mass = moles x MM = 1.0 mol Ag x 107.9 g/mol Ag = 107.9 g

Then, we calculate the heat required:

heat = mass x Cp x ΔT = 107.9 g x 0.24 J/°C ·g x 1.08°C = 27.96 J

C) Given:

heat = 1.25 kJ = 1,250 J

final T = 15.28°C

initial T = 12.08°C

We first calculate the change in temperature:

ΔT = final T - initial T = 15.28°C - 12.08°C = 3.2°C

Then, we calculate the mass of silver:

mass = heat/(Cp x ΔT) = 1,250 J/(0.24 J/°C ·g x 3.2°C) = 1,628 J ≅ 1.63 kJ