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Calculate the heat capacity of 25 grams of liquid ethanol at 25°C when it is heated from 25°C to 35°C by absorbing 150 Joules of heat. The specific heat of ethanol is 2.44 J/g°C.

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To calculate the heat capacity of the ethanol, we can use the formula:q = mcTwhere q is the heat absorbed  in Joules , m is the mass  in grams , c is the specific heat  in J/gC , and T is the change in temperature  in C .We are given the following information:- mass  m  = 25 grams- specific heat  c  = 2.44 J/gC- initial temperature  T1  = 25C- final temperature  T2  = 35C- heat absorbed  q  = 150 JoulesFirst, let's calculate the change in temperature  T :T = T2 - T1T = 35C - 25CT = 10CNow, we can plug the values into the formula:150 J =  25 g  2.44 J/gC  10C To solve for the heat capacity, we can divide both sides of the equation by the mass and the change in temperature:Heat capacity = 150 J /  25 g * 10C Heat capacity = 150 J / 250 gCHeat capacity = 0.6 J/gCThe heat capacity of the 25 grams of liquid ethanol at 25C when heated from 25C to 35C by absorbing 150 Joules of heat is 0.6 J/gC.

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