Question

1. A 56.6 -g piece of lead at 95.0 °C was immersed into 75.0 -g water at 25.0 °C. The water temperature rose to 26.6 °C. Calculate the molar heat capacity of the metal.
2. Calculate the energy needed to heat 12.1 g ice at -15.0 °C to liquid water at 70.0 °C. The heat of vaporization of water = 2257 J/g, the heat of fusion of water = 334 J/g, the specific heat capacity of water = 4.18 J/g·°C, and the specific heat capacity of ice = 2.06 J/g·°C.

Answer 1

(1)

Heat taken by water = heat released by metal

So,

$$m_wc_w(26.6-25)=m_mc_m(95-26.6)$$

$$75\times4.186\times 1.6=56.6\times c_m\times 68.4\\or\\c_m=0.13\ J\ gm^{-1}\ K^{-1}$$

Molar heat capacity = molar mass $\times\ c_m=26.73\ J\ mol^{-1}\ K^{-1}$

(2)

Total energy required = $m_ic_i(0-(-15))+(m_i\times334)+m_wc_w(70-0)$

$m_i=m_w=12.1\ g\\c_i=2.06\\c_w=4.18$

$E=(12.1\times2.06\times15)+(12.1\times334)+(12.1\times4.18\times70)\\\ \ \ \ =7955.75\ J$