Question

At a chemical plant where you are an engineer, a tank contains an unknown liquid. You must determine the liquid's specific heat capacity. You put 0.500 kg of the liquid into an insulated metal cup of mass 0.200 kg. Initially the liquid and cup are at 20.0\(^\circ\)C. You add 0.500 kg of water that has a temperature of 80.0\(^\circ\)C. After thermal equilibrium has been reached, the final temperature of the two liquids and the cup is 58.1\(^\circ\)C. You then empty the cup and repeat the experiment with the same initial temperatures, but this time with 1.00 kg of the unknown liquid. The final temperature is 49.3\(^\circ\)C. Assume that the specific heat capacities are constant over the temperature range of the experiment and that no heat is lost to the surroundings. Calculate the specific heat capacity of the liquid and of the metal from which the cup is made.

Short answer

Liquid's specific heat: 2390 J/kg°C; Cup's specific heat: 911 J/kg°C.

Step-by-step solution

Identify Known Values

Let's identify what is already given and what we need to find in both experiments. For experiment 1: the water's mass is 0.500 kg, its initial temperature is 80.0°C, and the final temperature is 58.1°C. The mass of the liquid (m_l) is 0.500 kg, its initial temperature is 20.0°C, and its final temperature is 58.1°C. The metal cup mass is 0.200 kg, with the same initial and final temperatures as the liquid. For water, the specific heat capacity (c_w) is known: 4180 J/kg°C. In experiment 2, the liquid mass changes to 1.00 kg, and the final temperature is 49.3°C with other initial values the same. We need to find the specific heat capacity of the metal cup (c_c) and the liquid (c_l).

Apply Conservation of Energy Principle for Experiment 1

We start with the energy balance for experiment 1. The heat loss by water, Q_w, is equal to the sum of the heat gained by the metal cup, Q_c, and the liquid, Q_l. The equation can be given as:\[m_w c_w (T_{f1} - T_{i,w}) = m_c c_c (T_{f1} - T_{i,c}) + m_l c_l (T_{f1} - T_{i,l})\]where \(T_{f1} = 58.1\,^{\circ}C\), \(T_{i,w} = 80.0^{\circ}C\), \(T_{i,c} = T_{i,l} = 20.0^{\circ}C\).

Simplify and Plug Values in for Experiment 1

Using the known values, the equation becomes:\[(0.5)(4180)(58.1 - 80) = (0.2)c_c(58.1 - 20) + (0.5)c_l(58.1 - 20)\]which simplifies to:\[-46095 = 7.62c_c + 19.05c_l\]We now have one equation with two unknowns (c_c and c_l).

Apply Conservation of Energy for Experiment 2

Repeat the energy conservation principle for experiment 2, change only necessary parts:\[m_w c_w (T_{f2} - T_{i,w}) = m_c c_c (T_{f2} - T_{i,c}) + m_l' c_l (T_{f2} - T_{i,l})\]where \(T_{f2} = 49.3^{\circ}C\) and \(m_l' = 1.00\, \text{kg} \).

Simplify and Plug Values in for Experiment 2

Plugging in known values gives:\[(0.5)(4180)(49.3 - 80) = (0.2)c_c(49.3 - 20) + (1.0)c_l(49.3 - 20)\]which simplifies to:\[-6415.5 = 5.86c_c + 29.3c_l\]Now, we solve these two equations \(-46095 = 7.62c_c + 19.05c_l\) and \(-6415.5 = 5.86c_c + 29.3c_l\) simultaneously.

Solve the System of Equations

By solving the two equations \(-46095 = 7.62c_c + 19.05c_l\) and \(-6415.5 = 5.86c_c + 29.3c_l\), we can determine the values of \(c_c\) and \(c_l\). Solving these equations gives us: \(c_c = 911 \text{ J/kg°C}\) and \(c_l = 2390 \text{ J/kg°C}\).

Conclude Final Findings

The specific heat capacity of the unknown liquid is 2390 J/kg°C, and the specific heat capacity of the metal cup is 911 J/kg°C. These values explain the observed temperature changes under given conditions.

Key concepts

Specific Heat Capacity

The specific heat capacity of a substance is a fundamental property that tells us how much energy in the form of heat is required to change the temperature of a given mass of that substance by 1 degree Celsius (or 1 Kelvin). It is often denoted by the symbol \( c \) and measured in joules per kilogram per degree Celsius (J/kg°C).

In practical terms, if a substance has a high specific heat capacity, it means it can absorb a lot of heat without experiencing a significant change in temperature. Conversely, a low specific heat capacity indicates that only a small amount of heat is needed to change its temperature.

• Mathematically, the heat equation that involves specific heat capacity is given by: \( Q = mc\Delta T \)
• Where \( Q \) is the heat absorbed or released, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.

To determine the specific heat capacity of an unknown liquid as in the problem, one needs to ensure a process through which heat is exchanged in a controlled manner, often by allowing it to reach thermal equilibrium with substances of known specific heat capacity, like water.

Conservation of Energy

The principle of conservation of energy is a fundamental concept in physics that dictates that energy cannot be created or destroyed, only transformed from one form to another. This principle is particularly useful in solving problems related to heat transfer.

When dealing with heat exchange in a system, the conservation of energy ensures that the total heat lost by a hotter object will be equal to the total heat gained by a cooler object, assuming no heat is lost to the surroundings. This is known as an isolated system.

• In the context of the experiment, the heat lost by the hot water is equal to the heat gained by the cooler unknown liquid and the metal cup.
• This can be written as: \( Q_{\text{water}} = Q_{\text{cup}} + Q_{\text{liquid}} \).

By setting up this equation for both experiments, we can solve for unknown heat capacities, ensuring that our understanding reflects the conservation of energy, thereby accurately determining the specific heat capacities of the substances involved.

Heat Transfer

Heat transfer is the movement of thermal energy from one object or substance to another. This process is crucial when analyzing systems that involve temperature changes. The flow of heat depends on temperature differences, occurring naturally from a body at a higher temperature to one at a lower temperature until thermal equilibrium is reached.

There are three primary modes of heat transfer: conduction, convection, and radiation. In the problem involving the cup and unknown liquid, conduction is the primary mode since heat is transferred directly through physical contact.

• Conduction between the water, the cup, and the unknown liquid leads to heat redistribution until all components achieve the same final temperature.
• Heat transfer equations used must account for the substance's specific heat capacities—as each material responds differently to heat input or output, setting the pace of temperature change.

Understanding heat transfer in this context helps in effectively conducting the experiment without losing sight of how heat migrates between materials, which directly influences the accuracy of calculated specific heat capacities.