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kiruha [24]
3 years ago
12

A 150g copper bowl contains 220g of water, both at 20.0oC, A very hot 300 g copper cylinder is dropped into the water, causing t

he water to boil, with 5.00 g being converted to steam. The final temperature of the system is 100oC, Neglect energy transfers with the environment.
a) How much energy (in calories) is transfered to the water as heat?
b) How much to the bowl?
c) What is the original temperature of the cylinder?
Physics
1 answer:
andrezito [222]3 years ago
3 0

Answer:

chk photo

Explanation:

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Batman (95kg) is standing on top of a 50m high building looking out over the city of Gotham. Given that he uses the potential en
Oksanka [162]

Answer:

47 kJoules (kJ)

Explanation:

Potential enegy on Earth is given by the relationship:

P.E. = mgh, where m is mass, g is the acceleration due to Earth's gravity, and h is height. Since we are given metric values, we will look for an answer that is consistent with Joules, the metric measure of energy. 1 Joule is defined as 1 kg*m^2/s^2, so we wnat units of kg, m, and sec.

We are given:

m = 95kg

h = 50 meters

Earth's gravity, g is 9.8 m/s^2

Enter the data:

P.E. = mgh

P.E. = (95kg)(9.8m/s^2)(50m)

P.E. = 46550 kg*m^2/s^2 or 46550 Joules(J)

Since we only have 2 sig figs, and since 1kJ =- 1000J

We can state the potential energy is 47kJ.

Spiderman has 47kJ of potential energy for the start of any dive back to Earth. [He needed that same amount of energy to reach that height, but we don't know from where it came. A jump, helicopter, beamed up by Scotty, or tossed up by Doctor Octopus.]

3 0
1 year ago
If your prediction turns out to be false your experiment has been a failure
nikdorinn [45]

This is false. Your hypothesis, or prediction, is just that: a prediction. Saying its a failure will result in bias.

3 0
4 years ago
Read 2 more answers
xConsider the following reduction potentials: Cu2+ + 2e– Cu E° = 0.339 V Pb2+ + 2e– Pb E° = –0.130 V For a galvanic cell employi
slega [8]

Answer:

Approximately \rm 90\; kJ.

Explanation:

Cathode is where reduction takes place and anode is where oxidation takes place. The potential of a electrochemical reaction (E^{\circ}(\text{cell})) is equal to

E^{\circ}(\text{cell}) = E^{\circ}(\text{cathode}) - E^{\circ}(\text{anode}).

There are two half-reactions in this question. \rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu and \rm Pb^{2+} + 2\,e^{-} \rightleftharpoons Pb. Either could be the cathode (while the other acts as the anode.) However, for the reaction to be spontaneous, the value of E^{\circ}(\text{cell}) should be positive.

In this case, E^{\circ}(\text{cell}) is positive only if \rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu is the reaction takes place at the cathode. The net reaction would be

\rm Cu^{2+} + Pb \to Cu + Pb^{2+}.

Its cell potential would be equal to 0.339 - (-0.130) = \rm 0.469\; V.

The maximum amount of electrical energy possible (under standard conditions) is equal to the free energy of this reaction:

\Delta G^{\circ} = n \cdot F \cdot E^{\circ} (\text{cell}),

where

  • n is the number moles of electrons transferred for each mole of the reaction. In this case the value of n is 2 as in the half-reactions.
  • F is Faraday's Constant (approximately 96485.33212\; \rm C \cdot mol^{-1}.)

\begin{aligned}\Delta G^{\circ} &= n \cdot F \cdot E^{\circ} (\text{cell})\cr &= 2\times 96485.33212 \times (0.339 - (-0.130)) \cr &\approx 9.0 \times 10^{4} \; \rm J \cr &= 90\; \rm kJ\end{aligned}.

5 0
3 years ago
Select the correct exercise category for each of the specific ways to increase intensity
Alecsey [184]
The answer is cardiovascular.
3 0
3 years ago
The Type K thermocouple has a sensitivity of about 41 uV /°C, i.e. for each degree difference in the junction temperature, the o
Novosadov [1.4K]

Answer:

ΔTmin = 3.72 °C

Explanation:

With a 16-bit ADC, you get a resolution of 2^{16}=65536 steps. This means that the ADC will divide the maximum 10V input into 65536 steps:

ΔVmin = 10V / 65536 = 152.59μV

Using the thermocouple sensitiviy we can calculate the smallest temperature change that 152.59μV represents on the ADC:

\Delta Tmin = \frac{\Delta Vmin}{41 \mu V/C}= 3.72 C

3 0
3 years ago
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