Please help! A student using the joulle heat apparatus records the following:?

Question

Mass aluminum cup = 80 grams
mass water = 375 grams
voltage= 6.0 V
current = 1.5 A
clock time= 17 min
initial water temp= 19.2 C
final water temp= 24.7 C
specific heat of water= 1.0cal/g C
specific heat aluminum= .215 cal/g C

1. what was the temp change of water (and aluminum cup)

2. how much heat did the water absorb?

3. how much heat did aluminum cub absorb?

4. how much energy was supplied to circuit?

5. based on results what value should be reported for the mechanicle equivalent of heat?

I was absent for this lesson please help!

Answer 1

Lets do some theory

Power delivered by the electric heater is
P=IV
I -current
V- voltage

Energy E= Pt
t - time
E(total)=Et=I V t=
Et= 1.5 x 6.0 x 17 x 60=9180 Joules
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It will help to know that 1 calorie = 4.187 Joules
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Since the system received energy hopefully it was all observed by the water and the container. That energy E in cal

E=9180 / 4.187 =2190 calories

E= E1 + E2 + E(loss) since E(loss) = 0
E= E1 + E2 =
E= Cp1m1(T2-T1) + Cp2m2(T2-T1)

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1) dT=T2-T1= 24.7C - 19.2 C=5.5 C

2) E(water)=Cp1 m1(T2-T1) =
E(water)= 1 x 375 x 5.5=2063 cal

3.)E(Al)=Cp2 m2(T2-T1) =
E(Al)= .215 x 80 x 5.5= 94.6 cal

4) E(supplies) = 2190 cal (computed earlier)

5) Friction work, mecahnical energy, is turned in to heat.

Just for curiosity

E(water )+E (Al)= 2063 + 94.6= 2158 cal

Loss = E(supplied) - E(obserbed)= 2190 - 2158 =32 cal