Answer:
1.06 °C
Explanation:
From the question given above, the following data were obtained:
Mass of gold (M₉) = 10 g
Specific heat capacity of gold (C₉) = 0.129 J/gºC
Initial temperature of gold (T₉) = 24 °C
Mass of water (Mᵥᵥ) = 118 g
Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gºC
Initial temperature of water (Tᵥᵥ) = 1 °C
Equilibrium temperature (Tₑ) =?
The equilibrium temperature of the system can be obtained as follow:
Heat loss by the gold = heat gained by the water
M₉C₉(T₉ – Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Cᵥᵥ)
10 × 0.129 (24 – Tₑ) = 118 × 4.184 (Tₑ – 1)
1.29(24 – Tₑ) = 493.712 (Tₑ – 1)
Clear bracket
30.96 – 1.29Tₑ = 493.712Tₑ – 493.712
Collect like terms
30.96 + 493.712 = 493.712Tₑ + 1.29Tₑ
524.672 = 495.002Tₑ
Divide both side by 495.002
Tₑ = 524.672 / 495.002
Tₑ = 1.06 °C
Therefore, the temperature of the system is 1.06 °C