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A medium-sized apple provides about 95.0 Calories of energy.

(a) Convert 95.0 Cal to joules.

(b) Suppose that amount of energy is transformed into kinetic energy of a 2.03 kg object initially at rest. Calculate the final speed of the object (in m/s).

(c) If that same amount of energy is added to 3.79 kg (about 1 gal) of water at 21.7°C, what is the water's final temperature (in °C)? The specific heat of water is

c = 4186

J

(kg · °C)

.

Respuesta :

Eduard22sly

A. 95 Cal is equivalent to 397480 J

B. The final speed of the object is 625.78 m/s

C. The final temperature of the water is 46.8 °C

A. Converting 95 Cal to Joules

1 Cal = 4184 J

Therefore,

95 Cal = 95 × 4184

95 Cal = 397480 J

Thus, 95 Cal is equivalent to 397480 J

B. Determination of the speed of the object

Kinetic energy (KE) = 397480 J

Mass (m) = 2.03 Kg

Velocity (v) =?

KE = ½mv²

397480 = ½ × 2.03 × v²

397480 = 1.015 × v²

Divide both side by 1.015

v² = 397480 / 1.015

Take the square root of both side

v = √(397480 / 1.015)

v = 625.78 m/s

Therefore, the speed of the object is 625.78 m/s

C. Determination of the final temperature of the water

Mass of water (M) = 3.79 kg

Initial temperature of water (T₁) = 21.7 °C

Specific heat capacity of water (C) = 4186 J/KgºC

Heat (Q) = 397480 J

Final temperature (T₂) =?

Q = MC(T₂ – T₁)

397480 = 3.79 × 4186 (T₂ – 21.7)

397480 = 15864.94 (T₂ – 21.7)

Clear the bracket

397480 = 15864.94T₂ – 344269.198

Collect like terms

397480 + 344269.198 = 15864.94T₂

741749.198 = 15864.94T₂

Divide both side by 15864.94

T₂ = 741749.198 / 15864.94

T₂ = 46.8 °C

Therefore, the final temperature of the water is 46.8 °C

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