**QUESTION 1**An ice cube is placed in an empty 80.0g aluminum cup (c_Al = 0.900 J/(g°C)). The ice cube (L_f = 333 J/g, L_v = 2260 J/g) has an initial temperature of 0.00°C while the cup has an initial temperature of 40.0°C. The ice cube completely melts, and the melted ice and the aluminum cup reach a final equilibrium temperature of 10.0°C. No heat is lost to the outside. Calculate the mass of the ice cube (in grams). ### Entropy Calculation Problems#### Question 4**Calculate the change in entropy for the aluminum cup (in J/K).***Input field for answer*#### Question 5**Calculate the total change in entropy for the entire system (in J/K).***Input field for answer*### Understanding Entropy ChangeEntropy is a measure of the disorder or randomness in a system. When heat is transferred within a system, the entropy of the system changes. This set of questions focuses on calculating these changes in entropy. For question 4, you need to calculate the change in entropy specifically for an aluminum cup. In question 5, you are required to calculate the total change in entropy for the entire system, which would include all components interacting thermally. To solve these questions, you will typically use the formula for the change in entropy (\(\Delta S\)):\[ \Delta S = \frac{Q}{T} \]Where:- \( \Delta S \) is the change in entropy- \( Q \) is the heat transferred- \( T \) is the absolute temperature in Kelvin at which the transfer takes placeFor scenarios involving different components or complex temperature changes, appropriate adjustments should be made based on the specifics of the situation and given data.

Given:- The mass of the aluminum cup is m =80g Specific heat of Aluminum is CA = 0.9J/g oC The ice…

n ice cube is placed in an empty 80.0g aluminum cup (CAI = 0.900J /(g °C )). he ice cube (Lf = 333] /g , Ly = 2260J /g ) has an initial temperature of 0.00°C while the cup has an initial temperature of 40.0°C . The ice cube completely nelts, and the melted ice and the aluminum cup reach a final equilibrium emperature of 10.0°C. No heat is lost to the outside. Calculate the mass of the ce cube (in grams).

n ice cube is placed in an empty 80.0g aluminum cup (CAI = 0.900J /(g °C )). he ice cube (Lf = 333] /g , Ly = 2260J /g ) has an initial temperature of 0.00°C while the cup has an initial temperature of 40.0°C . The ice cube completely nelts, and the melted ice and the aluminum cup reach a final equilibrium emperature of 10.0°C. No heat is lost to the outside. Calculate the mass of the ce cube (in grams).

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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