1.) An energy bill indicates that the customer used 953 kWh in July.How many jou
ID: 1001516 • Letter: 1
Question
1.) An energy bill indicates that the customer used 953 kWh in July.How many joules did the customer use?
E = J
2.) An adult eats food whose nutritional energy totals approximately 2100 Cal per day. The adult burns 1900 Cal per day.
a.) How much excess nutritional energy, in kilo joules, does the adult consume per day? Express your answer using one significant figure
E= kJ
b.) If 1 lb of fat is stored by the body for each 14.6×103kJ of excess nutritional energy consumed, how long will it take this person to gain 1 lb? Express your answer using one significant figure.
t= days
c.) Perform each of the following temperature conversions.
i.) 212 F to Celsius (temperature of boiling water) T= C
ii.) 77 K to Fahrenheit. (temperature of liquid nitrogen) Express your answer using two significant figures.
iii.) 25 C to kelvins (room temperature) T= k iv.) 98.9 F to kelvins (body temperature)
T = K
3). Vodka will not freeze in the freezer because it contains a high percentage of ethanol. The freezing point of pure ethanol is -114 C.
i.) Convert that temperature to degrees Fahrenheit
T= F
ii.) Convert that temperature to degrees Kelvin
T = k
4.) Calculate the amount of heat required to raise the temperature of an 84 g sample of water from 30 C to 64 C. Express your answer using two significant figures.
Q = J
Explanation / Answer
1.
Kw= Kj/sec
953 Kw= 953 Kj/sec
953 KwH= 953 Kj/sec* 3600 sec= 953*1000J/sec*3600 sec=3.431*109 J
2.
Excess fat= energy of the food he takes- energy burned= 2100-1900 =200 cal/day
1 Cal= 4.18 Joules. 200 Cal = 200*4.18 Joules /day =836 joules/ day= 0.836 Kj/day
3.
0.836 Kj correspond to 1 day
14.6*1000 Kj correspond to 14.6*1000/0.836 =17464 days
212= 1.8C+32
C= 100 deg.c
ii)
Deg.K= deg.c +273
77= deg.c +273
Deg.c= 77-273=-196
F= 1.8C+32= 1.8*(-196) +32 =-320.8 F
F= 1.8C+32
98.9= 1.8C+32
C= 37.2 deg.c = 37.2+273.15=310.35 K
F= 1.8C+32
F= 1.8*(-114)+32=-173.2 F
Deg. K= deg.C+273.15= -114+273.15=159.15K
Specific heat of water= 4.18 J/gm.deg.c
Heat required= mass* specific heat* temperature difference= 84*4.18*(64-30)=11938.08 Joules
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