Question

A piece of ice (heat capacity \(=2100 \mathrm{~J} \mathrm{~kg}^{-1}{ }^{\circ} \mathrm{C}^{-1}\) and latent heat \(=3.36 \times 10^{5} \mathrm{~J} \mathrm{~kg}^{-1}\) ) of mass \(\mathrm{m}\) grams is at \(-5^{\circ} \mathrm{C}\) at atmospheric pressure. It is given \(420 \mathrm{~J}\) of heat so that the ice starts melting. Finally when the ice-water mixture is in equilibrium, it is found that \(1 \mathrm{gm}\) of ice has melted. Assuming there is no other heat exchange in the process, the value of \(\mathrm{m}\) is

Short answer

Solving for m from the equation yields m = (420 J - 336 J) / (2100 J/kg°C × 5°C), which gives m = 0.016 kg or 16 grams.

Step-by-step solution

Determine the heat required to raise the temperature of ice to 0°C

First, let's calculate the heat required to raise the temperature of ice from -5°C to its melting point at 0°C using the formula: Q = mcΔT,where Q is the heat absorbed, m is the mass of ice, c is the specific heat capacity of ice, and ΔT is the change in temperature. Here, c = 2100 J/kg°C and ΔT = 5°C.

Determine the heat required to melt 1 gram of ice

The heat required to melt 1 gram of ice is given by the latent heat, which is 3.36×105 J/kg for ice. Since we have 1 gram of ice, we need to multiply this by the mass in kilograms (1 gram = 0.001 kg). So the heat required is Q = m_latent x L, where m_latent is the mass of ice that melted (1 gram) and L is the latent heat.

Calculate the total heat provided

Next, we need to calculate the total amount of heat that was provided, which is 420 J. This heat will be used both to raise the temperature of the ice to melting point and to melt 1 gram of it. So we can write the equation: Heat to raise temperature + Heat to melt ice = Total heat provided.

Construct the equation and solve for m

Now, we can set the sum of the heat required to raise the temperature of ice to 0°C and the heat required to melt 1 gram of ice equal to 420 J:m × 2100 J/kg°C × 5°C + 0.001 kg × 336000 J/kg = 420 J.We can now solve this equation for m.

Find the value of m

Solving the equation from step 4, we get:m × 10500 J/kg + 336 J = 420 J.Now, subtract 336 J from both sides, and then divide by 10500 J/kg to solve for m.

Key concepts

Specific Heat Capacity

Specific heat capacity is a fundamental concept in understanding how substances absorb and transfer heat. Imagine you are holding two different materials, a piece of metal and a block of wood, and you place them both in the sun. You'll notice the metal gets hot very quickly, but the wood takes more time to warm up. This behavior is due to their specific heat capacities.

Specific heat capacity () is defined as the amount of heat () required to raise the temperature of a unit mass of a substance by one degree Celsius (<>T>). Mathematically, the relationship is given by , where is the mass of the substance, and is the change in temperature. Materials with higher specific heat capacities require more energy to change their temperatures, making them useful in applications that require temperature regulation.

When heating our piece of ice from the exercise, understanding specific heat capacity helps us figure out how much energy we need to apply to raise its temperature to the melting point. This is crucial when dealing with processes involving temperature changes in substances.

Latent Heat of Fusion

The concept of latent heat of fusion comes into play when a material changes its phase, for example, from solid to liquid. As you've probably noticed, ice takes awhile to melt even after it's reached its melting point. This occurs because the ice needs extra energy for the molecular rearrangements that allow it to become water, which isn't accounted for by a temperature change.

This extra energy needed per unit mass to convert a solid into a liquid at a constant temperature (the melting point) is called the latent heat of fusion (). It's crucial to understand that during the phase change, all the energy supplied as heat is used for changing the state, without raising the temperature.

The exercise requires that we calculate the energy to melt exactly 1 gram of ice. Here, the latent heat of fusion becomes vital: it allows us to determine the amount of energy required for the phase transition, calculated as , where is the mass that has undergone the phase change. It's this understanding that enables us to account for those molecules that have made the leap from solid to liquid, which is a separate consideration from simply warming up the material.

Thermodynamics

Thermodynamics is a branch of physics that studies the movement and conversion of energy, particularly heat, and how it affects matter. It's based on four fundamental laws, which apply to the exercise at hand. The first law of thermodynamics, also known as the law of conservation of energy, states that energy cannot be created or destroyed, only transformed from one form to another.

In our case, the heat energy provided is used to do two things: raise the temperature of ice and convert some of the ice into water (melting). This is a practical example of the first law – the conservation of energy principle. We use the equations for specific heat and latent heat to calculate the total energy involved in these processes. By ensuring the total heat provided equals the sum of heat used to warm the ice and the heat needed to melt it, we're applying the principles of thermodynamics to solve a real-world problem.

Energy Conversion

Energy conversion is essentially walking hand in hand with the first law of thermodynamics. It refers to changing energy from one form to another. This concept is pivotal not only in physics but also in our everyday lives; it's how we power our world, from car engines to cooking our food.

In our exercise, when heat is provided to the system (the ice), energy conversion happens: the thermal energy from a hotter object (perhaps a stove or the surrounding air) is transferred to the ice, increasing its internal energy and enabling the temperature rise or the phase change from solid to liquid.

Acknowledging this energy conversion helps us to understand that the provided 420 J is not lost, but instead it transforms the state and temperature of the ice. Correctly accounting for energy conversion is key for tasks in thermodynamics, and particularly so in solving the given exercise where precise energy accounting leads to determining the unknown mass of ice, .