Question

A house with a well pump gets groundwater at a temperature of 50°F. The 55-gallon hot water heater is set to maintain water at 120°F. A typical "hot" shower is 100°F - not much above skin temperature. If the hot water heater isn't heating the water (maybe due to a power outage), but the tank is full of 120°F water, and if a shower consumes 3.0 gallons of water per minute, how long will a hot shower last? Please write this out like a mixing chamber problem in the chapter and solve.

          A house with a well pump gets groundwater at a temperature of 50°F. The 55-gallon hot water heater is set to maintain water at 120°F. A typical "hot" shower is 100°F - not much above skin temperature. If the hot water heater isn't heating the water (maybe due to a power outage), but the tank is full of 120°F water, and if a shower consumes 3.0 gallons of water per minute, how long will a hot shower last? Please write this out like a mixing chamber problem in the chapter and solve.
        
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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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A house with a well pump gets groundwater at a temperature of 50°F. The 55-gallon hot water heater is set to maintain water at 120°F. A typical "hot" shower is 100°F - not much above skin temperature. If the hot water heater isn't heating the water (maybe due to a power outage), but the tank is full of 120°F water, and if a shower consumes 3.0 gallons of water per minute, how long will a hot shower last? Please write this out like a mixing chamber problem in the chapter and solve.
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Transcript

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00:01 We'd like to find how much time is required to raise the temperature of the glass.
00:07 So let's go ahead and first find our change in temperature.
00:11 So our temperature is changing from 53 degrees from 20 to 53 degrees, so that's going to be 33 degrees celsius.
00:20 Now we know that our heat is equal to mass times our specific heat times change in temperature.
00:27 So we can write q is equal to the mass of gallon in our tank, so that's going to be our 50 gallons.
00:35 So that's 189 .25 liters times 4 .186 times our change in temperature which is 33 degrees celsius.
00:50 So let's go ahead and plug that into a calculator.
00:53 189 .25 times 4 .186 times 33, we get 26142 .6.
01:02 Now our required time t is our q over w.
01:07 So that's our 26142 .6 over our w, where our w is our power or 5650 watts...
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