Acurrent of 0,5a flows through a 22 resistor for 100s. the system who- se initial temperature is 300k is thermally isolated. the heat capacity of the resistor is 0,24jk-1 for a wide range of temperatures. the temperatu- re of the resistor changes appreciably. what is the entropy change of the system?

The entropy change of the system is 0.1265 J/K.

Explanation:

Given that,

Current = 0.5 A

Resistor = 2 Ω

Time = 100 s

Temperature = 300 K

Heat capacity = 0.24 J/K

We need to calculate the heat produced

Using formula of heat

Where, i = current

R = resistor

t = time

Put the value into the formula

We need to calculate the change in temperature

Using formula of temperature

Now, The temperature is

We need to calculate the entropy change of the system

Using formula of entropy

Hence, The entropy change of the system is 0.1265 J/K.

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form