A 10.0-g ice cube at -10.0^∘ C is dropped into 40.0 g of water at 30.0^∘ C. a) After enough tim

A 10.0-g ice cube at -10.0^∘ C is dropped into 40.0 g of...

Key Concepts

Specific Heat Capacity

This concept refers to the amount of energy required to raise the temperature of a unit mass of a substance by one degree Celsius. It is crucial when calculating the sensible heat changes in substances, meaning the energy absorbed or released while the substance's temperature changes without a phase transition.

Latent Heat

Latent heat is the energy absorbed or released by a substance during a phase change at a constant temperature. It is essential for analyzing processes like melting or freezing, where energy is used to change the phase of the material rather than changing its temperature.

Conservation of Energy

This key concept states that in an isolated system, the total energy remains constant. When different substances at varying temperatures interact, the heat lost by the warmer substance must equal the heat gained by the cooler one, making this principle central to solving equilibrium problems.

Thermal Equilibrium

Thermal equilibrium is the state reached when all parts of a system have the same temperature, with no net heat flow between them. Understanding this concept is important when determining the final temperature after different temperature bodies are mixed or interact over time.

Phase Change Processes

This encompasses the phenomena associated with substances transitioning from one phase to another, such as melting or freezing. It involves latent heat and is significant in problems where a substance undergoes a change of state while interacting with other substances at different temperatures.

Transcript

00:01 So this problem, one has to determine the equilibrium temperature for this two situations.

00:09 First, we have added one ice cube and second, if two ice cube is added.

00:18 Just given in the problem with the initial mass of the water is 40 grams, and the initial temperature of water is 30 degrees celsius, and the mass of the ice cube, these 10 grams and the initial temperature of the ice is negative 10 degrees celsius.

00:50 Now, if not all ice melt, the final temperature or the equilibrium temperature is equals to 0 degrees.

01:08 But if not, if all ice melt, the equilibrium temperature is not equal to 0 degrees.

01:19 So let's determine situation a, which is one ice cube is added, if in this case all i smelt or not all ice melt.

01:45 So all aisles milk if the absolute value of the heat required to change the temperature of the water to the melting point of the ice, which is 0 degree celsius, is greater than or equal to the sum, the heat required to change the temperature of the water to the melting point of the ice, which is 0 degree celsius, is greater than or equal to the sum.

02:44 Required to change the temperature ice to the 3 degrees celsius and the heat required to change the face of ice, which is to change ice from solid, which is to liquid.

03:30 Getting this heat, the numerical value of this heat, then we have first for the heat required to change the temperature 40 to the melting point of ice.

03:40 This is our equation.

03:46 Mass of the water, multiply the specific heat of the water, with the temperature difference, which is 0 degrees celsius minus initial temperature of water.

03:56 So we have 0 .04 kilogram multiplied by 4190 ,000, 0 .190, 0 .000, 0 .000 ,000, 0 .000 ,000 ,000 ,000 ,000 ,000.

04:09 And the answer is negative 5 ,028 shows.

04:15 For the heat required for the ice to change the temperature to the melting point.

04:21 So, the ratio and mass of the ice, multiplied by the specific heat of the ice, multiplied by the changing temperature, which is 0 degrees minus the initial temperature of the ice.

04:33 So we have 0 .1 kilogram, 3 .01, we convert the grams to kilogram, multiplied by 220, dose per kilogram kelvin, which is a specific heat of ice in may.

04:50 The changing temperature 0 degrees minus negative 10 degrees celsius we can find this is equals to 206 joules now for the heat required to change the face of the ice then we have its mass 0 point mass of ice multiplied by the latin heat of fusion of water so we have 0 .01 kilogram multiplied by 3 .34 times 10 raised to 5 jous per kilogram and we can find this is equals to 3 ,340 jules.

05:29 Now, let us determine if all ice will melt or not.

05:33 Then we have, so this is the case when one ice cube is added.

05:57 We have absolute value of negative 5 ,028 jules.

06:04 Is it greater than or equal to the sum of 260 joles plus 3 ,3 ,040 jules? we can find that.

06:19 It's indeed greater than greater than 300, 3 ,546 jokes.

06:39 Therefore, in a situation when one ice cube is added, all ice melt.

06:47 So the equilibrium temperature is not equals to zero.

06:51 So we will determine through this.

06:56 As all ice melt, it will reach equilibrium temperature where we have this heat, heat of the melted ice required to change its temperature to the equilibrium temperature, which is equal to the mass of the ice...

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