Calculate the number of joules given off when 32.0 grams of steam cools from 110.0 °C to ice at -40.0 °C.
First the steam should cool to 100°C, next it should turn into water and cool to 0°C, then the water should turn into ice and the ice should cool to -40.0°C. Let us write the amount of energy corresponding to each stage of process.
1) "Q_1 = c_s m_s \\Delta T_s = 2078\\cdot 0.032\\cdot 10\u00b0\\mathrm{C} = 665\\,\\mathrm{J}"
2) "Q_2 = r m_s = 2.26\\cdot10^6\\cdot0.032= 72320\\,\\mathrm{J}"
3) "Q_3 = c_w m_w \\Delta T_w = 4200\\cdot0.032\\cdot100\u00b0\\mathrm{C} = 13440\\,\\mathrm{J}"
4) "Q_4 = \\lambda m_w = 330000\\cdot0.032 = 10560\\,\\mathrm{J}"
5) "Q_5 = c_im_i\\Delta T_i = 2110\\cdot0.032\\cdot40\u00b0\\mathrm{C} = 2701\\,\\mathrm{J}."
The total amount of energy will be "665+72320+13440+10560+2701 =99686\\,\\mathrm{J}\\approx 10^5\\,\\mathrm{J}."
Comments
Leave a comment