How much heat (in kj) is required to warm 11.0 g of ice, initially at -13.0 ∘c, to steam at 111.0 ∘c? the heat capacity of ice is 2.09 j/g⋅∘c and that of steam is 2.01 j/g⋅∘c?

The total heat required will be 33.35 KJ.

Explanation:

In this question there are two kinds of processes.

1) Heat required for change in temperate = mC∆T

2) Heat required for state change = mass × heat of fusion or latent heat of vaporization

We need some other constants to solve this question.

Heat of fusion of water = 334 J/g

Latent heat of vaporization = 2230 J/g

Specific heat of water = 4.184 J/g0C  

Heat required to change temperature from -13 to 0 degree centigrade

                                Q1 = (11)(2.09)(13) = 298.87 J

Heat required in converting ice to water at 0 degree centigrade

                                Q2 = (11)(334) = 3674 J

Heat required to change temperature from 0 to 100 degree centigrade

                                Q3 = (11)(4.184)(100) = 4602.4 J

Heat required in converting water to water vapors at 100 degree centigrade                  

                                 Q4 = (11)(2230) = 24530 J

Heat required to change temperature from 100 to 111 degree centigrade

                                  Q5 = (11)(2.01)(11) = 243.21 J

Total heat required will be:

                                  Q = Q1 + Q2 + Q3 + Q4 + Q5

                         Q  = 298.87 + 3674 + 4602.4 + 24530 + 243.21 = 33348.48 J

                                   Q = 33.35 KJ

answer: question1

1,000 tons

question2

3: 8

question 3

store d

step-by-step explanation:

it is given that trillions of microbes make up to 1% to 3% of the body mass.

let's take average of this percentage: 2%

consider the average mass of the human body as 70 kg.

then, trillions of microbes would have mass  

therefore, 100 trillion micro-organisms would weigh:

this is an approximate value and could change depending on the mass of the human body and percentage of the microbes present in it.

average mass per microorganism =