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Chapter 2: Problem 14

A 12.9-g sample of an unknown metal at 26.5° is placed in a Styrofoam cup containing 50 g of water at 88.6°. The water cools down, and the metal warms up until thermal equilibrium is achieved at 87.1°. Assuming all the heat released by the water is absorbed by the metal and that the cup is perfectly insulated, determine the specific heat of the unknown metal. The specific heat of water is 4,186 J/(kg·K). A. 1,401.03 J/(kg·K) B. 401.03 J/(kg·K) C. 4,010.3 J/(kg·K) D. 40,103 J/(kg·K)

Short Answer

Expert verified
The specific heat of the unknown metal is 401.03 J/(kg·K), which corresponds to option B.

Step by step solution

01

Identify given values

Given values:- Mass of the metal (\(m_{\text{metal}}): 12.9 g = 0.0129 kg- Initial temperature of the metal (\)T_{\text{initial, metal}}): 26.5°C- Mass of the water (\(m_{\text{water}}): 50 g = 0.05 kg- Initial temperature of the water (\)T_{\text{initial, water}}): 88.6°C- Final temperature (\(T_{\text{final}}): 87.1°C- Specific heat of water (\)c_{\text{water}}): 4,186 J/(kg·K)
02

Calculate the heat lost by the water

Use the formula:\[ q_{\text{water}} = m_{\text{water}} \times c_{\text{water}} \times \Delta T_{\text{water}} \]where \(\Delta T_{\text{water}} = T_{\text{initial, water}} - T_{\text{final}} \).\[ \Delta T_{\text{water}} = 88.6 - 87.1 = 1.5 \text{°C} \]\[ q_{\text{water}} = 0.05 \ kg \times 4186 \ \frac{J}{kg\cdot K} \times 1.5 \text{ K} = 313.95 \text{ J} \]
03

Assume heat gained by the metal equals heat lost by the water

Since all the heat lost by the water is absorbed by the metal, we have:\[ q_{\text{metal}} = q_{\text{water}} \]Thus, \( q_{\text{metal}} = 313.95 \text{ J} \).
04

Calculate the specific heat of the metal

Use the formula:\[ q_{\text{metal}} = m_{\text{metal}} \times c_{\text{metal}} \times \Delta T_{\text{metal}} \]where \(\Delta T_{\text{metal}} = T_{\text{final}} - T_{\text{initial, metal}} \).\[ \Delta T_{\text{metal}} = 87.1 - 26.5 = 60.6 \text{°C} \]Now solve for \(c_{\text{metal}}\):\[ 313.95 \text{ J} = 0.0129 \text{ kg} \times c_{\text{metal}} \times 60.6 \text{ K} \]\[ c_{\text{metal}} = \frac{313.95 \text{ J}}{0.0129 \text{ kg} \times 60.6 \text{ K}} \approx 401.03 \ \frac{J}{kg\cdot K} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

heat transfer
Heat transfer is the process by which heat energy moves from one object to another. This occurs due to a temperature difference between the objects. In the context of the provided exercise, heat was transferred from the hot water to the cooler metal. This transfer continued until both substances reached the same temperature, known as thermal equilibrium. Heat transfer can take place through various methods such as conduction (direct contact), convection (fluid movement), and radiation (emission of energy).
thermal equilibrium
Thermal equilibrium is a state reached when two bodies in thermal contact no longer exchange heat energy. In the exercise, the water and metal reached thermal equilibrium when both stabilized at a final temperature of 87.1°C. This state means that the energy gained by the metal was exactly equal to the energy lost by the water. The concept is crucial to understanding how heat transfer calculations are done. It helps in determining the specific heat capacity of substances involved by analyzing the heat exchange process.
specific heat capacity
The specific heat capacity is a property that describes how much heat energy is required to raise the temperature of 1 kilogram of a substance by 1 K (or °C). This is different for different materials. In the exercise, the specific heat capacity of the unknown metal was calculated using the heat transfer formula: \[ q = m \times c \times \triangle T \] where:
  • \(q\) is the heat transferred
  • \(m\) is the mass of the substance
  • \(c\) is the specific heat capacity
  • \(\triangle T\) is the temperature change
By knowing the amount of heat absorbed or released and the resultant temperature change, we can solve for the specific heat capacity.
temperature change
Temperature change (\( \triangle T \) ) is the difference between the final temperature and the initial temperature of a substance. It plays a crucial role in calculating heat transfer. In the given exercise, the temperature changes for both water and metal were necessary to determine the specific heat capacity of the unknown metal. For water: \[ \triangle T_{\text{water}} = T_{\text{initial, water}} - T_{\text{final}} \] and for the metal: \[ \triangle T_{\text{metal}} = T_{\text{final}} - T_{\text{initial, metal}} \] These changes in temperature helped in applying the heat transfer equations correctly and led to the final calculation.

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Most popular questions from this chapter

Hydroelectric dams make use of the differences in water levels to generate electricity. Which of the following shows the energy conversion in a dam? A. Gravitational potential energy \(\rightarrow\) Kinetic energy \(\rightarrow\) Electrical energy B. Electrical energy \(\rightarrow\) Heat energy \(\rightarrow\) Electrical energy

What is the kinetic energy, in joules, of a \(625-\mathrm{kg}\) roller coaster car that is moving with a speed of \(18.3 \mathrm{~m} / \mathrm{s}\) ? A. \(1.05 \times 10^4 \mathrm{~J}\) B. \(1.05 \times 10^5 \mathrm{~J}\) C. \(2.05 \times 10^5 \mathrm{~J}\) D. \(1.06 \times 10^4 \mathrm{~J}\)

A system has \(60 \mathrm{~J}\) of heat added to it, resulting in \(15 \mathrm{~J}\) of work being done by the system and exhausting the remaining \(45 \mathrm{~J}\) of heat. What is the efficiency of this process? A. \(100 \%\) B. \(60 \%\) C. \(45 \%\) D. \(25 \%\)

Which of the following is NOT a property of a gas? A. no definite shape B. can be compressed C. definite volume D. definite mass

At room temperature, which of the following liquids has the highest density? A. orange juice B. milk C. honey D. water
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