(a)If a metal block(initially at room temperature, which we assume is 22°C in th

Question

(a)If a metal block(initially at room temperature, which we assume is 22°C in this case) with a mass of 150g were dropped from a height of 6 m into 250g of water (also initially at room temperature) in a container, what would be the change in the temperature of the water?

For simplicity, assume in this case that the container does not absorb any heat. Use the conversion relation 1 cal = 4.186 Joule in your calculations.

(b)

List the various kinds of energies that are

Involved in this interaction(e.g. heat, kinetic energy, etcetera).

(c)

What caused the change in the water temperature? Explain your reasoning.

Explanation / Answer

a)

Potential energy lost by the metal throught the height h = 6m,

PE = mgh = 0.15*9.8*6 = 8.82 J

Now, this energy is used to heat up the water,

So, 8.82 = M*C*dT

where M = 0.25 kg

C = 4186 J/kg/C

dT = change in temperature of water = T - 22

So, 0.25*4186*(T - 22) = 8.82

So, T = 22.0084 deg C

So, change in temperaure = 0.0084 deg C

b)

Firstly Potential energy is converted to Kinetic energy then this kinetic energy is converted to heat of the water

c)

The friction due to the motion of the metal inside the water caused the metal to lose energy and thus this energy is converted to heat energy for increasing the temperature of water

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