Q: A 92.9 g piece of a silvery gray metal is heated to 178.0∘C, and then quickly transferred into 75.0 mL of water initially at 24.0∘C. After 5 minutes, both the metal and the water have reached the same temperature: 29.7∘C. Determine the specific heat and the identity of the metal. You are given that water=4.184 J/g.°C, and the density of water is 1.00 g/mL.
Answer:
0.1297 J/g.°C
The metal is lead.
Explanation:
Assuming no heat is lost to the surrounding,
Heat lost by the gray metal = heat gained by water
c'm'(t₁-t₃) = cm(t₃-t₁)......................... Equation 1
Where, c' = specific heat capacity of the gray metal, m' = mass of the gray metal, c = specific heat capacity of water, m = mass of water, t₁ = initial temperature of the gray metal, t₂ = initial temperature of water, t₃ = temperature of the mixture,
Making c' the subject of the equation,
c' = cm(t₃-t₁)/m'(t₁-t₃) ...................... Equation 2
Given: m' = 92.9 g, c = 4.184 J/g.°C, t₁ =178°C, t₂ =24°C, t₃ = 29.7°C
But m = density of water× volume
m = 1.00×75 = 75 g
Substitute into equation 2,
c' = 4.18×75(29.7-24)/[92.9(178-29.7)]
c' = 1786.95/13777.07
c' = 0.1297 J/g.°C
Hence The specific heat capacity of the metal = 0.1297 J/g.°C
The specific heat capacity of the metal is approximately equal to that of lead which is 0.13 J/g.°C. Hence the metal is lead.
The heat capacity of the silvery gray metal is 0.131 J/g∘C.
Volume of water = 75 mL
Mass of water (m2) =75 g
mass of metal(m1) = 92.9 g
Final temperature of the system (T) = 29.7∘C
Initial temperature of metal (Tm) = 178.0∘C
Initial temperature of water (Tw) = 24.0∘C
Heat capacity of water (cw) = 4.184 J/g∘C
Heat capacity of metal (cm) = ?
Heat lost by metal = Heat gained by water
Heat lost by metal = m1cm(Tm - T)
Heat gained by water = m2cw(T - Tw)
Hence;
m1cm(Tm - T) = m2cw(T - Tw)
92.9 × cm × ( 178.0 - 29.7) = 75 × 4.184 × (29.7 - 24.0)
cm = 75 × 4.184 × (29.7 - 24.0)/92.9 × ( 178.0 - 29.7)
cm = 0.131 J/g∘C
Learn more: https://brainly.com/question/9352088
