An iron rod of mass 2 kg and at a temperature of 280°C is dropped into some quantity of water initially at a temperature of 30°C. If the temperature of the mixture is 70°C, calculate the mass of the water. (Neglect heat losses to the surroundings) [Specific heat capacity of iron = 460 Jkg\(^{-1}\)K\(^{-1}\)] [Specific heat capacity of water = 4200 Jkg\(^{-1}\)K\(^{-1}\)]
A
1.50 kg
B
0.58 kg
C
1.15 kg
D
0.77 kg
Verified Correct Answer: C.
Step-by-Step Mathematical Solution
m\(_s\) = 2 kg; c\(_s\) = 460 Jkg\(^{-1}\)K\(^{-1}\); m\(_w\) = ?; c\(_w\) = 4200 Jkg\(^{-1}\)K\(^{-1}\); θ\(_1\) = 280°C; θ\(_2\) = 30°C; θ\(_3\) = 70°C
Heat loss = Heat gained
m\(_s\) c\(_s\) (θ\(_1\) - θ\(_3\)) = m\(_w\) c\(_w\) (θ\(_3\) - θ\(_2\))
2 × 460 × (280 - 70) = m\(_w\) × 4200 × (70 - 30)
2 × 460 × 210 = m\(_w\) × 4200 × 40
193,200 = 168,000 m_w
m\(_w\) = \(\frac{193200}{168000}\) ≈ 1.15kg.
Source: WAEC / JAMB / NECO Classroom Archive (Record #76360)