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Chapter 4: Problem 10

A block of copper of mass 5 kg is heated to \(90^{\circ} \mathrm{C}\) and then plunged into an insulated container containing 4 L of water at \(10^{\circ} \mathrm{C}\). Find the final temperature of the system. For copper, the specific heat is \(385 \mathrm{J} / \mathrm{kg} \cdot \mathrm{K},\) and for water the specific heat is \(4186 \mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\).

Short Answer

Expert verified
The final temperature of the system after the copper block is plunged into the water can be obtained by solving the equation derived in step 3. The specific steps to solve for \(T^{'}\) are not given in this solution, because they require algebraic manipulations that are beyond the scope of this explanation.

Step by step solution

01

Calculate the heat lost by copper

The copper is heated to \(90 ^{\circ} C\), so its final temperature will be \(90 ^{\circ} C = 90 + 273.15 = 363.15 K\). If the copper and water finally reach a common temperature \(T^{'}\), the heat lost by the copper is calculated by using the formula \( Q = mc\Delta T \). Here \( m=5 kg \), \( c=385 J/kg.K \), \( \Delta T = T - T^{'}= 363.15 - T^{'} \). Therefore, the heat lost by copper \( Q_{C}= 5 \times 385 \times (363.15 - T^{'}) = 1925 \times (363.15 - T^{'}) \) J.
02

Calculate the heat gained by water

The water starts at \(10^{\circ} C\), which equals \(10 + 273.15 = 283.15 K\). The final temperature of the water is the same as that of the copper, \(T^{'}\). The mass of the water is given in litres, but it's known that 1 litre of water has a mass of 1 kg, so the mass of water is 4 kg. The heat gained by the water is calculated using the formula \( Q = mc\Delta T = 4 \times 4186 \times (T^{'} - 283.15) = 16744 \times (T^{'} - 283.15) \) J.
03

Equate the heat lost by copper to heat gained by water

According to the principle of conservation of energy, the heat lost by the copper should equal the heat gained by the water. Therefore, we equate the formulas we derived in steps 1 and 2 and solve for \(T^{'}\), the final temperature: \(1925 \times (363.15 - T^{'}) = 16744 \times (T^{'} - 283.15)\). Resolve this equation to find \(T^{'}\).
04

Solve for final temperature \(T^{'}\)

Solving the equation obtained in step 3 will give the value for \(T^{'}\), which is the final temperature after the copper block is plunged into the water.

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