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Chapter 14: Problem 86

A blacksmith heats a 0.38 -kg piece of iron to \(498^{\circ} \mathrm{C}\) in his forge. After shaping it into a decorative design, he places it into a bucket of water to cool. If the available water is at \(20.0^{\circ} \mathrm{C},\) what minimum amount of water must be in the bucket to cool the iron to \(23.0^{\circ} \mathrm{C} ?\) The water in the bucket should remain in the liquid phase.

Short Answer

Expert verified
Answer: The minimum amount of water needed to cool the iron to 23.0°C is approximately 12.71 kg.

Step by step solution

01

Identifying the Given Information

We are given the following information: - Mass of iron (\(m_1\)) = 0.38 kg - Initial temperature of iron (\(T_{i1}\)) = \(498^{\circ} \mathrm{C}\) - Final temperature of iron (\(T_{f1}\)) = \(23^{\circ} \mathrm{C}\) - Initial temperature of water (\(T_{i2}\)) = \(20^{\circ} \mathrm{C}\) - Final temperature of water (\(T_{f2}\)) = \(23^{\circ} \mathrm{C}\) We need to find the minimum amount of water (mass of water, \(m_2\)). The specific heat capacities of iron and water are: - Specific heat capacity of iron (\(c_1\))= 450 J/kg\(\cdot\)K - Specific heat capacity of water (\(c_2\))= 4190 J/kg\(\cdot\)K
02

Writing the Heat Transfer Equation

Since heat lost by iron is equal to heat gained by water, we can write the equation: \(m_1 \cdot c_1 \cdot (T_{f1} - T_{i1}) = m_2 \cdot c_2 \cdot (T_{f2} - T_{i2})\)
03

Calculating the Change in Temperature

First, let's find the change in temperature for both iron and water: - Change in temperature of iron (\(\Delta T_1\)) = \(T_{f1} - T_{i1}\) = \(23^{\circ} \mathrm{C} - 498^{\circ} \mathrm{C} = -475^{\circ} \mathrm{C}\) - Change in temperature of water (\(\Delta T_2\)) = \(T_{f2} - T_{i2}\) = \(23^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C} = 3^{\circ} \mathrm{C}\)
04

Plugging Values into the Heat Transfer Equation

Now we can plug all the values into the heat transfer equation: \((0.38\,\text{kg}) \cdot (450\,\text{J/kg}\cdot\text{K}) \cdot(-475\,\text{K}) = m_2 \cdot (4190\,\text{J/kg}\cdot\text{K}) \cdot (3\,\text{K})\)
05

Solving for the Minimum Mass of Water (\(m_2\))

Solve the equation to find the minimum mass of water: \(m_2 = \dfrac{(0.38\,\text{kg}) \cdot (450\,\text{J/kg}\cdot\text{K}) \cdot(-475\,\text{K})}{(4190\,\text{J/kg}\cdot\text{K}) \cdot (3\,\text{K})}\) \(m_2 \approx 12.71\,\text{kg}\) Therefore, the minimum amount of water needed to cool the iron to \(23.0^{\circ} \mathrm{C}\) is approximately 12.71 kg.

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Most popular questions from this chapter

A metal rod with a diameter of \(2.30 \mathrm{cm}\) and length of $1.10 \mathrm{m}\( has one end immersed in ice at \)32.0^{\circ} \mathrm{F}$ and the other end in boiling water at \(212^{\circ} \mathrm{F}\). If the ice melts at a rate of 1.32 g every 175 s, what is the thermal conductivity of this metal? Identify the metal. Assume there is no heat lost to the surrounding air.
How much heat is required to change \(1.0 \mathrm{kg}\) of ice, originally at \(-20.0^{\circ} \mathrm{C},\) into steam at \(110.0^{\circ} \mathrm{C} ?\) Assume 1.0 atm of pressure.
A \(0.400-\mathrm{kg}\) aluminum teakettle contains \(2.00 \mathrm{kg}\) of water at \(15.0^{\circ} \mathrm{C} .\) How much heat is required to raise the temperature of the water (and kettle) to \(100.0^{\circ} \mathrm{C} ?\)
Imagine that 501 people are present in a movie theater of volume $8.00 \times 10^{3} \mathrm{m}^{3}$ that is sealed shut so no air can escape. Each person gives off heat at an average rate of \(110 \mathrm{W} .\) By how much will the temperature of the air have increased during a 2.0 -h movie? The initial pressure is \(1.01 \times 10^{5} \mathrm{Pa}\) and the initial temperature is \(20.0^{\circ} \mathrm{C} .\) Assume that all the heat output of the people goes into heating the air (a diatomic gas).
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