Question

A piece of copper of mass 300g at a temperature of 950 Celsius is quickly transferred to a vessel of negligible thermal capacity containing 250g of water at 25 Celsius . If final steady temperature of the mixture is 100 ceslsius.calculate the mass of the water that boil away . (Specific heat capacity of copper=4.2×10^2. specific heat capacity of water=4.2×10^3. latent heat of vaporization of steam=2.26×10^6)

Answer 1

$$m_cc_c(T_c-T)=m_wc_w(T-T_w)+mL$$

$$m=\frac{(300)(420)(950-100)-(250)(4200)(100-25)}{2.26\cdot 10^6}$$

$$m=12.5\ g$$