Question

A 70 -lbm copper block initially at \(220^{\circ} \mathrm{F}\) is dropped into an insulated tank that contains \(1.2 \mathrm{ft}^{3}\) of water at \(65^{\circ} \mathrm{F}\) Determine \((a)\) the final equilibrium temperature and \((b)\) the work potential wasted during this process. Assume the surroundings to be at \(65^{\circ} \mathrm{F}\).

Short answer

In the scenario where a 70 lbm copper block at an initial temperature of 220°F is dropped into a 1.2 ft³ tank of water with an initial temperature of 65°F, the final equilibrium temperature of the system is found to be 77.42°F. Additionally, the work potential wasted during this process is calculated to be 2226.18 Btu.

Step-by-step solution

Calculate the mass of water and copper block

First, convert the given volume of water to mass, using the density of water, which is commonly taken as \(62.4 \,\mathrm{lbm/ft^3}\). Also, note down the mass of the copper block. Water: Density \(\rho_{w} = 62.4 \mathrm{\frac{lbm}{ft^3}}\) Volume \(V_{w} = 1.2 \mathrm{ft^3}\) \(M_{w} = \rho_{w} \times V_{w} = 62.4\,\mathrm{\frac{lbm}{ft^3}}\times 1.2 \,\mathrm{ft^3} = 74.88\, \mathrm{lbm}\) Copper Block: Mass \(M_{c} = 70\, \mathrm{lbm}\)

Find specific heat capacities

Look up the specific heats for water and copper, usually given in Btu/(lbm°F): Water: \(c_{w} = 1 \mathrm{\frac{Btu}{lbm \cdot ^{\circ}F}}\) Copper: \(c_{c} = 0.092 \mathrm{\frac{Btu}{lbm \cdot ^{\circ}F}}\)

Apply energy balance equation

Assuming no heat is lost to surroundings, the heat lost by the copper block must equal the heat gained by the water: \(M_{c}c_{c}\left(T_{f} - T_{c_i}\right) = M_{w}c_{w}\left(T_{w_i} - T_{f}\right)\) Where \(T_{f}\) is the final equilibrium temperature, \(T_{c_i} = 220^{\circ} \mathrm{F}\) is the initial temperature of the copper block and \(T_{w_i} = 65^{\circ} \mathrm{F}\) is the initial temperature of the water. Plugging in the known values, we have: \(70\, \mathrm{lbm} \times 0.092\, \mathrm{\frac{Btu}{lbm \cdot ^{\circ}F}} \times (T_{f} - 220^{\circ} \mathrm{F}) = 74.88\, \mathrm{lbm} \times 1\, \mathrm{\frac{Btu}{lbm \cdot ^{\circ}F}} \times (65^{\circ} \mathrm{F} - T_{f})\)

Solve for final temperature

Solve the energy balance equation for \(T_{f}\): \(6.44\, \mathrm{Btu/^{\circ}F}\times (T_{f} - 220^{\circ} \mathrm{F}) = 74.88\, \mathrm{Btu/^{\circ}F} \times (65^{\circ} \mathrm{F} - T_{f})\) Solving for \(T_{f}\) gives: \(T_{f} = 77.42^{\circ} \mathrm{F}\) Thus, the final equilibrium temperature is \(77.42^{\circ} \mathrm{F}\). #b. Finding the Work Potential Wasted#

Calculate the change in entropy

We will first calculate the change in entropy for the copper block and the water separately, and then find the total change in entropy in the system: Entropy change for the copper block: \(\Delta S_{c} = M_{c}c_{c}\ln{\frac{T_{f}}{T_{c_i}}}\) Entropy change for the water: \(\Delta S_{w} = M_{w}c_{w}\ln{\frac{T_{f}}{T_{w_i}}}\) Now let's plug the values: \(\Delta S_{c} = 70\, \mathrm{lbm} \times 0.092\, \mathrm{\frac{Btu}{lbm \cdot ^{\circ}F}} \times \ln{\frac{77.42}{220}} = -6.889\, \mathrm{Btu\cdot ^{\circ}F}\) \(\Delta S_{w} = 74.88\, \mathrm{lbm} \times 1\, \mathrm{\frac{Btu}{lbm \cdot ^{\circ}F}} \times \ln{\frac{77.42}{65}} = 2.324\, \mathrm{Btu\cdot ^{\circ}F}\) Total change in entropy: \(\Delta S = \Delta S_{c} + \Delta S_{w} = -6.889 + 2.324 = -4.565\, \mathrm{Btu\cdot ^{\circ}F}\)

Calculate the work potential wasted

The work potential wasted can be calculated as: \(W_{wasted} = -T_{surr} \times \Delta S\) As the surroundings are at \(65^{\circ} \mathrm{F}\), we first convert it to Rankine scale and use \(T_{surr}=493.67\, \mathrm{R}\) \(W_{wasted} = -\left(493.67\, \mathrm{R}\right) \times (-4.565\, \mathrm{Btu\cdot ^{\circ}F}) = 2226.18\, \mathrm{Btu}\) Thus, the work potential wasted during this process is \(2226.18\, \mathrm{Btu}\).