Heat transfer and calorimetry

[boomboompoop]

id appreciate some input u guys...

Heat Transfer and Calorimetry

A 50 g Aluminum calorimeter contains 95 g of a mixture of water and ice at 0 degree Celsius. When 100 g of Aluminum, which has been heated in a steam jacket to 100 degree Celsius, is dropped into the mixture, the temperature rises to 5 degree Celsius. Find the mass of ice originally present in the mixture if the specific heat capacity of aluminum is 0.22 cal/g*C

 

[nocturnal]

How about you show your work. If you are completely lost on how to begin this problem, and thus have no work to show, then you should look up the heat/temperature change equations and the heat/phase change equations. Also remember the conservation of energy laws - heat gained equals heat lossed.

 

[kyle8921]

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Thank you for sharing this interesting problem. I would approach this situation by first understanding the principles of heat transfer and calorimetry. Heat transfer refers to the movement of thermal energy from one object to another, while calorimetry is the measurement of heat transfer during a physical or chemical process. In this scenario, we are dealing with both heat transfer and calorimetry as the hot aluminum is dropped into the mixture of ice and water, resulting in a change in temperature.

To solve this problem, we can use the equation Q = mcΔT, where Q is the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

First, we can calculate the heat transferred from the hot aluminum to the mixture of ice and water. We know that the initial temperature of the mixture is 0 degrees Celsius, and after the aluminum is added, the temperature rises to 5 degrees Celsius. Therefore, the change in temperature is ΔT = 5 - 0 = 5 degrees Celsius.

Next, we can calculate the heat transferred using the given mass and specific heat capacity of aluminum. Using the equation Q = mcΔT, we get Q = 100 g x 0.22 cal/g*C x 5 degrees Celsius = 110 cal.

Since the heat transferred from the aluminum is equal to the heat absorbed by the ice and water, we can set up an equation to solve for the mass of ice, denoted as m1, in the initial mixture. This can be written as:

110 cal = (m1 + 95 g) x 1 cal/g*C x (5 - 0) degrees Celsius

Solving for m1, we get m1 = 15 g. Therefore, the initial mass of ice in the mixture was 15 g.

In conclusion, by understanding the principles of heat transfer and calorimetry and using the appropriate equations, we were able to determine the mass of ice in the initial mixture. This problem showcases the practical application of these concepts in real-life scenarios.