What mass of water will change its temperature by 3.00 °C when 595 J of heat is added to it? The specific heat of water is 4.2 J
/g °C
1 answer:
M = 36.5g
^ degrees sign
Q energy heat in JOULES J
m mass of the sample in GRAMS g
C specific heat in J/g^C
ΔT change in temperature ^C
or final temp- initial temp
Based on the information you've provided, we know the following variables:
ΔT= 3^C
Q = 525 J
C= 4.8 J
g×^C
All we have to do is rearrange the equation to solve for m. This can be accomplished by dividing both sides by C and ΔT to get m by itself like this:
m= Q
ΔT×C
Now, we just plug in the known values:
m= 525 J
3^C×4.8 J
g×^C
m=36.5g
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True is the correct answer! your welcome! :)
Given:
volume of 0.08 m³
density of 7,840 kg/m³
Required:
force of gravity
Solution:
Find the mass using density
equation.
D = M/V
M = DV
M = (7,840 kg/m³)(0.08 m³)
M = 627.2kg
F = Mg
F = (627.2kg)(9.8m/s2)
F = 6147N
D. Ap3x just did it
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<h3>
Answer:</h3>
d) 1 and 4
<h3>Explanation:</h3>
- Elements are made up of particles known as atoms.
- Atoms of the same element have the same properties as the element.
- The atoms of an element may differ in terms of the number of neutrons and have a different mass number, such atoms are known as isotopes.
- However, the isotopes have an equal number of protons or atomic number.
- In this case, atom 1 has a mass number of 20 and an atomic number of 10 while atom 4 has a mass number of 21 and an atomic number of 21.
- Therefore, atom 1 and 4 are isotopes of element E.
