Why cant you put a metals in microwaves, and toaster, etc <br><br> i have been told but never taught

What mass of LP gas is necessary to heat 1.4 L of water from room temperature (25.0 ∘C) to boiling (100.0 ∘C)? Assume that durin

DochEvi [55]

Answer:

$m_{LP}=0.45\,kg$

Explanation:

Let assume that heating and boiling process occurs under an athmospheric pressure of 101.325 kPa. The heat needed to boil water is:

$Q_{water} = (1.4\,L)\cdot(\frac{1\,m^{3}}{1000\,L} )\cdot (1000\,\frac{kg}{m^{3}} )\cdot [(4.187\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (100^{\textdegree}C-25^{\textdegree}C)+2257\,\frac{kJ}{kg}]$

$Q_{water} = 3599.435\,kJ$

The heat liberated by the LP gas is:

$Q_{LP} = \frac{3599.435\,kJ}{0.16}$

$Q_{LP} = 22496.469\,kJ$

A kilogram of LP gas has a minimum combustion power of $50028\,kJ$ . Then, the required mass is:

$m_{LP} = \frac{22496.469\,kJ}{50028\,\frac{kJ}{kg} }$

$m_{LP}=0.45\,kg$

6 0

3 years ago

Suppose we use radix sort to sort the English-language strings below using standard lexicographic ordering (i.e. sort in alphabe

nlexa [21]

Answer:

a) 4 passes are required to sort the string.

b) 4

c) i) TARP

ii) CHIP

iii) PART

iv) TARP

v) TARP

d) O(k+n), n is no. of strings, k is largest no. of character in among the string

O(d*(n+10)), n is no. of integers

Explanation:

7 0

3 years ago

2. What are the separate parts that make up an oxyacetylene flame?

amm1812

Answer:

Explanation:

Primary combustion where carbon monoxide (CO) and free hydrogen (H) are produced, and the secondary combustion where water vapor (H2O) and carbon dioxide (CO2) are formed.

6 0

3 years ago

Consider a pond that initially contains 10 million gallons of fresh water. Water containing a chemical pollutant flows into the

Ostrovityanka [42]

Answer / Explanation:

To solve this , we start by recalling the conversion principle:

Thus, dc/dt = Rate of C in - Rate of C out,

Therefore, for the number of chemical C(t) , i gram, in the pond at time (t), the rate of flow of C is modeled as:

Rate of Cin = Cin F,

Where Cin is the concentration of the chemical going in ( in this case, Cin = 2 grams per gallon), and F is the rate of flow of water into the pond (in this case, F = 5 million gallons per year). We then have that

Rate of C in = 10 million grams per year

Therefore the rate of Cout = c(t)F, where c(t) = C(t) / V is the concentration of the chemical in the pond ( here, we are assuming instant mixing). The volume, (V) of the water in the pond is 10 million gallons (we are assuming that the rate of flow of water into the pond is the same as the rate of flow out, so that the volume of water in the pond remains constant). Thus,

Rate of C out = 1/2 C(t) million grams per year.

Thus, the differential equation describing the evolution of C =

C(t) in time is: dc/dt = 10 - C/2 in millions of grams per year.

Now to plot the solution using Maple and to describe in words the effect of the variation in the incoming chemical,

we have: C(t) = 20 +ce ∧ -t/2, where c is an arbitrary constant since there was no chemical in the pond initially.

Therefore, C(t) = 0 , then C = -20

So that Q(t) = 20 ( 1 - e ∧ -t/2)

Kindly find the expression of the above statement C = C(t) below:

Note: In the diagram, Q is represented by C

Therefore, C = Q