Question

A blacksmith dropped a 1.5 kg iron bead at 300oC into some quantity of water. If the temperature of the water rose from 15oC to 18oC, what is the mass of the water assuming no heat is lost to the surrounding? (Take the specific heat of iron as 0.46 J kg-1 C-1 and that of water as 4.2x103 J kg-1 C-1)
A. 15.44 kg B. 194.58 g C. 15.44 g
D. 194.58 kg

Answer 1

From the conservation of energy:

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

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

$$m_w=15\ kg$$