Question

A 2.0 kg ice cube at 0 oC is placed in 22 kg of water at 25 oC. Determine the final

temperature of the mixture. [Given the specific heat of water, cw is 4186 J/kg oC and the

latent heat of fusion of water, Lf is 3.33 ×105 J/kg]

Answer 1

$qm_1+c_wm_1(t-0°)=c_wm_2(25°-t)\to$

$t=\frac{c_wm_2\cdot25°-qm_1}{c_w(m_1+m_2)}=\frac{4186\cdot 22\cdot25°-330000\cdot 2}{4186\cdot (2+22)}=16.3°C$ . Answer