Question

A blacksmith cools a 1.30 kg chunk of iron, initially at a temperature of 650.0∘C, by trickling 15.4 ∘C water over it. All the water boils away, and the iron ends up at a temperature of 120.0∘C.

How much water did the blacksmith trickle over the iron?

Express your answer with the appropriate units.

Answer 1

Solution :

Given :

mass of iron (m_iron) = 1.3 kg

Initial temperature of iron (T_oi) = 650.0^o C

Initial temperature of water (T_ow) = 15.4^o C

Final temperature (T_f) = 120.0^o C

According to conservation of energy :

Therefore: Heat lost by iron = Heat gained by water

m_iron C_iron (T_oi - T_f) = m_water { C_water (100 - T_ow) + C_vapour (T_f - 100) + L_vapour }

(1.3) (0.47) (650 - 120) = m_water { (4.186)(100 - 15.4) + (1.996)(120 - 100) + (2260) }

323.83 = m_water { 2654.0556 }

m_water= 0.122 kg