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Digiron [165]
2 years ago
10

In a much smaller model of the Gizmo apparatus, a 5 kg mass drops 86 mm (0.086 m) and raises the temperature of 1 gram of water

1°C. How much energy was transferred to the water? (Use a value of 9.8 m/s2 for g.)
Engineering
1 answer:
Orlov [11]2 years ago
6 0

Answer:

The amount of energy transferred to the water is 4.214 J

Explanation:

The given parameters are;

The mass of the object that drops = 5 kg

The height from which it drops = 86 mm (0.086 m)

The potential energy P.E. is given by the following formula

P.E = m·g·h

Where;

m = The mass of the object = 5 kg

g = The acceleration de to gravity = 9.8 m/s²

h = The height from which the object is dropped = 0.086 m

Therefore;

P.E. = 5 kg × 9.8 m/s² × 0.086 m = 4.214 J

Given that the potential energy is converted into heat energy, that raises the 1 g of water by 1°C, we have;

The amount of energy transferred to the water = The potential energy, P.E. = 4.214 J.

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Explanation:

<u>Given  </u>

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-The volume of the tank V_tank = 1 m^3  

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-The increase in specific internal energy Δu = +10 kJ/kg  

-The change in kinetic energy ΔKE = 0 and The change in potential energy  

ΔPE =0  

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(a)Specific volume at the final state v_2

(b)The energy transferred by the work W in kJ.  

(c)The energy transferred by the heat transfer W in kJ and the direction of  

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Assumption  

-Quasi-equilibrium process.  

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