Question

Electrical Energy objective Measure the specific heat of water using an electric heating element АС Power Supply VRMs = 6.30 Volts Electric power P = VI Pava = VRMS IRMS Thermal power Q = mc AT dQ = mc dT w pode - cmwcw+m ca de Calorimeter Water

Answer 1

Suppose water and Calorimeter are at temperature mass of Calorimeter = mo mass of water amw. specific a heat of water = Cw Specific heat of Calorimeter a la when Current flowing through the resistance of a for a time of interval dt. power dissipation in resistor - I'R which increases the temperature of Calorimeter and water to the heat gain by the Calorimeter and water = m . Cc (T.-T2) +mw Cw (Ti-T2) da. (melt mwaw) at TAT=Tit2. me tmww) dT = IR DE ma w dT + M, GAT = R dt a mo Gw dt = IR dt- meldT - I'R dt - me le dT I modt. 16.30 R dt - meedT I = (²) mada - 6.30 (6.30 dt - med T Ma dT.