Question

A blacksmith dropped a 1.5kg iron bead at 300°C into some quantity of water. If the temperature of the water rose from 15°C to 180°C, calculate the mass of the water. (specific heat of iron = 0.46J/kg/C) (water= 4.2*10^3J/kj/C)

Answer 1

From the conservation of energy:

$$m_i c_i (T_i-T)=m_w c_w (T-T_w)$$

$$(1.5)(460)(300-18)=m_w (4200) (18-15)$$

$$m_w=15 kg$$