Question

You are part of an engineering firm on contract by the U.S. Department of​ Energy's Energy Efficiency and Renewable Energy task force to develop a program to help consumers measure the efficiency of their home appliances. Your job is to measure the efficiency of​ stove-top burners. The consumer will place a pan of room temperature water on their stove​ (with 11 gallon of​ water), record the initial room temperature in units of degrees​ Fahrenheit, turn on the​ burner, and wait for it to boil. When the water begins to​ boil, they will record the time in units of minutes it takes for the water to boil.​ Finally, they will look up the power for the burner provided by the manufacturer.Using all of this information provided by the​ user, determine the efficiency of their burner. Assume the specific heat of water is 4.18 joules per gram degree Celsius

Answer 1

V = 11 gal --> 3.78 L/gal = 11*3.78 = 41.6395 liters

D = 1 L/kg = 41.6395 L --> 41.6395 kg fo water = 41.6395*10^3 g = 41639.5 g

If Water is boiling --> 100°C

At T = 25°C

dT = 100-25 = 75 °C

C = 4.184 J/gC

Q = m*C*dT = (41639.5¿)(4.184)(75) = 13066475.1 J

assume

Energy inlet = x

then

Efficiency = Q/x * 100% = 13066475.1 / x *100%

Note that there is no data for Energy inlet so assume:

15000000 J inlet

so

Efficiency = 13066475.1 /15000000 *100 = 87.109834 % approx