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]
"qm_1+c_wm_1(t-0\u00b0)=c_wm_2(25\u00b0-t)\\to"
"t=\\frac{c_wm_2\\cdot25\u00b0-qm_1}{c_w(m_1+m_2)}=\\frac{4186\\cdot 22\\cdot25\u00b0-330000\\cdot 2}{4186\\cdot (2+22)}=16.3\u00b0C" . Answer
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