Answer:
26.7°C
Explanation:
Using the formula; Q = m × c × ΔT
Where; Q = amount of heat
m = mass
c = specific heat
ΔT = change in temperature
In this question involving iron placed into water, the Qwater = Qiron
For water; m= 50g, c = 4.18 J/g°C, Initial temp= 25°C, final temp=?
For iron; m = 30.5g, c = 0.449J/g°C, Initial temp= 52.7°C, final temp=?
Qwater = -(Qiron)
m × c × ΔT (water) =- {m × c × ΔT (iron)}
50 × 4.18 × (T - 25) = - {30.5 × 0.449 × (T - 52.7)}
209 (T - 25) = - {13.6945 (T - 52.7)}
209T - 5225 = -13.6945T + 721.7
209T + 13.6945T = 5225 + 721.7
222.6945T = 5946.7
T = 5946.7/222.6945
T = 26.7
Hence, the final temperature of water and iron is 26.7°C
