a) Convert the following energy units. a) 755 kJ to J? b) 2.90 × 103 kcal to kJ?

Question

a) Convert the following energy units. a) 755 kJ to J?

b) 2.90 × 103 kcal to kJ?

c) 6.82 × 106 J to kcal

-A 28.00 g sample of a substance is initially at 28.8 °C. After absorbing 668 cal of heat, the temperature of the substance is 103.5 °C. What is the specific heat (SH) of the substance?

SH=?? cal/g x degrees Celcius

-The specific heat of a certain type of cooking oil is 1.75 cal/(g·°C). How much heat energy is needed to raise the temperature of 2.81 kg of this oil from 23 °C to 191 °C?

=?? cal

Explanation / Answer

1)

a)

1 kj = 1000 j

so 755 KJ = 755*1000 = 755,000 J

b)

1 KCal = 4.186 KJ

so 2.90 × 103 KCal = 2.90*103*4.186 = 1250.3582KJ

c)

6.86*106 J = 6.86*106 /1000 KJ = 0.72716 KJ

Now 1 KCal = 4.186 KJ

so 1 KJ = 1/4.186 KCal

    0.72716KJ = 0.72716*1/4.186 KCal    = 0.1737KCal

2)

m = 28.00 g

T1 = 28.8 °C

T2 = 103.5 °C

T2-T1 = 103.5- 28.8 = 74.7

q = 668 cals

Now q= m*SH*(T2-T1)

   668 = 28*SH*(74.7)

so SH = 668/28*74.7 = 0.3194 cal/g x degrees Celcius

3)

SH = 1.75 cal/g x degrees Celcius

m = 2.81 kg = 2.81*1000 g

T2-T1 = 191-23 = 168°C

q = m*SH*(T2-T1)

= 2.81*1000*1.75*168 = 826140 cal

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.