Calculate the energy required to boil the mass m of water. The electrical energy "P\\tau" from the heater will heat up the vessel (heat required is "C\\Delta t") and the water (heat required "cm\\Delta t"):
"Q=cm\\Delta t+C\\Delta t=P\\tau,\\\\\ncm\\Delta t+C\\Delta t=\\frac{V^2}{R}\\tau,\\\\\\space\\\\\nm=\\frac{1}{c}\\bigg(\\frac{V^2\\tau}{R\\Delta t}-C\\bigg)=\\\\\\space\\\\\n=\\frac{1}{4200}\\bigg(\\frac{220^2\\cdot2\\cdot60}{20(100-40)}-100\\bigg)=1.13\\text{ kg}."
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