Question

A physics student uses a 115.00 V immersion heater to heat 330.00 grams of water for herbal tea. During the two minutes it takes the water to heat, the physics student becomes bored and decides to figure out the resistance of the heater. The student starts with the assumption that the water is initially at the temperature of the room T_i = 25.00°C and reaches T_f = 100.00°C.The specific heat of the water is c = 4180 J/(kg · °C). What is the resistance (in Ω) of the heater?

_____________________Ω

Answer 1

30 slag © here we have, Ti = 95.0% Tf - 100.0% specific heat of Wale = c = 4180 J/kg Dessipated heat by heater = &- pt - pr Power of heater to time taken. And heat absorbed by water to Rise tempratue to at a & -mc of - ma mass ca specific heat $T "Change in temprature So, combining 0 and ③ we have pa meot of ( If a PC) - (loo - 29750 m → 330 g = 330 kg t q min = 2x60 = 120 seconds. So, P = 330 x 4180 x 7500 - z 862.125 W 720 .

h ar And Paul R = x LY = tery p=862.15w So, K = (1155? - [15.340 ] 862.125