Question

A mass of \(1.4 \mathrm{kg}\) of water at \(22^{\circ} \mathrm{C}\) is poured from a height of \(2.5 \mathrm{m}\) into a vessel containing \(5.0 \mathrm{kg}\) of water at \(22^{\circ} \mathrm{C} .\) (a) How much does the internal energy of the \(6.4 \mathrm{kg}\) of water increase? (b) Is it likely that the water temperature increases? Explain.

Short answer

Short Answer: The internal energy of the water increases by 34.3 J after pouring 1.4 kg of water from a height of 2.5 m into a vessel containing 5.0 kg of water. However, the temperature of the water is not likely to noticeably increase, as the estimated temperature increase is only around 0.0020 K.

Step-by-step solution

(a) Calculate the gravitational potential energy (GPE) of the 1.4 kg water at the height of 2.5 m

To find the GPE of the 1.4 kg water, we can use the following formula: GPE = m * g * h, where m is the mass of water, g is the gravitational acceleration (approximately 9.8 m/s²), and h is the height from which the water is poured. Plug in the given values and calculate the GPE: GPE = 1.4 kg * 9.8 m/s² * 2.5 m.

(a) Calculate the increase in internal energy

The increase in internal energy is equal to the change in gravitational potential energy when the water is poured into the vessel. Since all of the potential energy is transferred into the internal energy of the water in the vessel, we have: Internal energy increase = GPE

(b) Analyze whether the water temperature increases or not

The specific heat capacity of water is approximately 4186 J/kg·K. Since the internal energy increase is relatively small compared to the total heat capacity of 6.4 kg of water, it is not likely that the temperature of the water will increase significantly. The energy is rapidly dispersed into the large mass of water, which acts as a buffer, preventing a temperature change. Now apply the formulas from the outlined steps above: Solution:

(a) Calculate the gravitational potential energy (GPE) of the 1.4 kg water at the height of 2.5 m

GPE = (1.4 kg) * (9.8 m/s²) * (2.5 m) = 34.3 J

(a) Calculate the increase in internal energy

Internal energy increase = GPE = 34.3 J

(b) Analyze whether the water temperature increases or not

Since the internal energy increase is 34.3 J and the specific heat capacity of water is 4186 J/kg·K, the temperature increase can be estimated using the formula: ∆T = Q / (m * c), where ∆T is the temperature change, Q is the internal energy increase, m is the total mass of water, and c is the specific heat capacity of water. ∆T = 34.3 J / (6.4 kg * 4186 J/kg·K) ≈ 0.0020 K. As the temperature increase is very small, it is not likely that the water temperature will noticeably increase.