Answer :
m₁ = mass of water = 75 g
T₁ = initial temperature of water = 23.1 °C
c₁ = specific heat of water = 4.186 J/g°C
m₂ = mass of limestone = 62.6 g
T₂ = initial temperature of limestone = ?
c₂ = specific heat of limestone = 0.921 J/g°C
T = equilibrium temperature = 51.9 °C
using conservation of heat
Heat lost by limestone = heat gained by water
m₂c₂(T₂ - T) = m₁c₁(T - T₁)
inserting the values
(62.6) (0.921) (T₂ - 51.9) = (75) (4.186) (51.9 - 23.1)
T₂ = 208.73 °C
in three significant figures
T₂ = 209 °C
Final answer:
To find the specific heat capacity of the metal, we can use the formula: q = mcΔT, where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Substituting the given values and solving, we find that the specific heat capacity of the metal is approximately 0.252 J/g °C.
Explanation:
To find the specific heat capacity of the metal, we can use the formula: q = mcΔT, where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. In this case, we know the initial and final temperatures of the metal, as well as the mass of the water, so we can plug these values into the equation. Rearranging the equation to solve for c, we have: c = q / (mΔT). Substituting the values and solving, we find that the specific heat capacity of the metal is approximately 0.252 J/g °C.
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