Answer:
1092 J
Explanation:
From the question,
Assuming no heat is lost to the surrounding
Heat lost by the metal = Heat gained by water.
H' = Cm(t₂-t₁)................. Equation 1
Where H' = Heat lost by the metal, C = specific heat capacity of water, m = mass of water, t₂ = Final temperature of the mixture, t₁ = Initial temperature of water
But,
m = D'v................... Equation 2
Where D' = Density of water, v = volume of water.
Given: D' = 1000 kg/m³, v = 100 mL = 100/1000000 = 0.0001 m²
Substitute into equation 2
m = 1000(0.0001)
m = 0.1 kg.
Also given: C = 4200 J/kg.°C, t₁ = 22 °C, 24.6 °C
Substitute into equation 1
H' = 4200×0.1×(24.6-22)
H' = 420(2.6)
H' = 1092 J.
Hence the heat (q) lost by the metal = 1092 J
