An immersion heater must increase the temperature of 1.40 kg of water from 10.0 C to 50.0 C in 10.0 min while operating at 110 V. (A) What is the required resistance of the heater? (B) Estimate the cost of heating the water. SOLVE IT (A) What is the required resistance of the heater? Conceptualize An immersion heater is a resistor that is inserted into a container of water. As energy is delivered to the immersion heater, raising its temperature, energy leaves the surface of the resistor by heat, going into the water. When the immersion heater reaches a constant temperature, the rate of energy delivered to the resistance by electrical transmission is equal to the rate of energy delivered by heat to the water. Categorize This example allows us to link our new understanding of power in electricity with our experience with specific heat in thermodynamics. The water is a nonisolated system. Its internal energy is rising because of energy transferred into the water by heat from the resistor: Aint = Q. In our model, we assume the energy that enters the water from the heater remains in the water. Analyze To simplify the analysis, let's ignore the initial period during which the temperature of the resistor ncreases and also ignore any variation of resistance with temperature. Therefore, we imagine a constant rate of energy transfer for the entire 10.0 min. Set the rate of energy delivered to the P = (AV2 - resistor equal to the rate of energy Q entering the water by heat Use the equation, Q = mcAT, to relate the = mcal - R = (AV)-At energy input by heat to the resulting At mcAT temperature change of the water and solve for the resistance: Substitute the values given in the R =. (110 V)2(600 s) statement of the problem: (1.40 kg) (4186 ]/kg . "C) (50.0 C - 10.0.C) R = 30.97 Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. B) Estimate the cost of heating the water. Multiply the power by the time interval to 1 h ind the amount of energy transferred: TET = PAt = (AV At = (110 V)2 (10.0 min) "(60.0 min) TET = x kwh Find the cost knowing that energy is Cost = (TT)($0.11/kWh) = x purchased at an estimated price of 11c per kilowatt-hour: Finalize The cost to heat the water is very low, less than one cent. In reality, the cost is higher because some energy is transferred from the water into the surroundings by heat and electromagnetic radiation while its temperature is increasing. If you have electrical devices in your home with power ratings on them, use this power rating and an approximate time interval of use to estimate the cost for one use of the device. MASTER IT HINTS: I'M STUCK! What is the required resistance of an immersion heater that will melt 6.06 kg of ice at 0C in 35 min while operating at 110 V? (Assume the latent heat of fusion of water is 3.33 x 10 J/kg.) 2 =