Water of volume 2 L in a closed container is heated with a coil of 1 kW. While water is heated, the container loses energy at a rate of 160 J/s. In how much time will the temperature of water rise from 27 o C to 77 o C ? (Specific heat of water is 4.2 kJ/kg and that of the container is negligible).

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Aicharya · Accepted Answer

P = 1 0 0 0 &#160;watt &#8756;&#160;&#160;&#160; P in = 1 0 0 0 - 1 6 0 &#160; J/s = 8 4 0 &#160;J/s Now, d Q dt = ms dT dt &#8756;&#160;&#160;&#160;&#160; 8 4 0 = ms dT dt &#8756;&#160;&#160;&#160; dt = ms 8 4 0 dT Integrating both sides we have; t = ms 8 4 0 &#8747; dT = 2 &#215; 1 0 0 0 &#215; 1 0 - 6 &#215; 1 0 0 0 &#215; 4.2 &#215; 1 0 0 0 8 4 0 &#215; 5 0 = 5 0 0 sec = 8&#160; min 2 0 sec

Water of volume 2 L in a closed container is heated with a coil of 1 kW. While water is heated, the…

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