A balloon is cooled by removing 0.784 kJ of heat. It shrinks on cooling, and the atmosphere does 655 J of work on the balloon. Calculate the change in internal energy, ∆E, of the system.

The correct answer is -0.129 kJ

Explanation:

In the given case, the cooling of the balloon is done by withdrawing 0.784 J of heat, and the work done by the atmosphere on the balloon is 655 J. First, there is a need to transform kJ into J, 1 kJ = 1000 J. So, 0.784 kJ would be 784 J.  

The ΔE or the change in the internal energy can be calculated by using the formula, ΔE = q + w (1).  

In the given case, q refers to the heat moved out of the system, that is, the value of q would be less than 0 or will be a negative quantity. Therefore, the heat moved out of the system will be -784 J. On the other hand, as the work or w is done on the system, therefore, the value of w would be more than 0 or will be a positive quantity. Thus, the value of w will be +655 J.  

Now putting the values in the equation (1) we get,  

ΔE = -784 J + 655 J

ΔE = -129 J or -0.129 kJ

As the change in internal energy comes out to be a negative value, therefore, the process is considered exothermic.  

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally