Answer:
For iron
Final temperature = 54,22°C
For copper
Final Temperature = 63.67 °C
Explanation
Hello,
You are using a torch to warm up a block of iron that has an initial temperature of 32°C.
The first you have to know is that the "heat capacity" could simply define as the heat required to go from an initial temperature to a final temperature.
So you need to use the heat capacity equation as follow in the paper.
The equation has to have all terms in the same units, so:
q = 12000 J
s = 0.450 J / g °C
m = 1200 g
Ti = 32 °C
Phosphoryl chloride :
POCl3
<span>
hope this helps!.
</span>
This is a straightforward question related to the surface energy of the droplet.
<span>You know the surface area of a sphere is 4π r² and its volume is (4/3) π r³. </span>
<span>With a diameter of 1.4 mm you have an original droplet with a radius of 0.7 mm so the surface area is roughly 6.16 mm² (0.00000616 m²) and the volume is roughly 1.438 mm³. </span>
<span>The total surface energy of the original droplet is 0.00000616 * 72 ~ 0.00044 mJ </span>
<span>The five smaller droplets need to have the same volume as the original. Therefore </span>
<span>5 V = 1.438 mm³ so the volume of one of the smaller spheres is 1.438/5 = 0.287 mm³. </span>
<span>Since this smaller volume still has the volume (4/3) π r³ then r = cube_root(0.287/(4/3) π) = cube_root(4.39) = 0.4 mm. </span>
<span>Each of the smaller droplets has a surface area of 4π r² = 2 mm² or 0.0000002 m². </span>
<span>The surface energy of the 5 smaller droplets is then 5 * 0.000002 * 72.0 = 0.00072 mJ </span>
<span>From this radius the surface energy of all smaller droplets is 0.00072 and the difference in energy is 0.00072- 0.00044 mJ = 0.00028 mJ. </span>
<span>Therefore you need roughly 0.00028 mJ or 0.28 µJ of energy to change a spherical droplet of water of diameter 1.4 mm into 5 identical smaller droplets. </span>
.150 = 3
20.20 = 4
25.00 = 4
If a 0 does not have a number or a period after it, it is not significant.
If the 0 is behind the decimal point, it is always significant.