Answer:
recall that heat absorbed released is given by
Q = mc*(T2 - T1)
where
m = mass (in g)
c = specific heat capacity (in J/g-k)
T = temperature (in C or K)
*note: Q is (+) when heat is absorbed and (-) when heat is released.
substituting,
Q = (480)*(0.97)*(234 - 22)
Q = 98707 J = 98.7 kJ
Explanation:
Answer:
Aluminium
Explanation:
When a body is immersed in a liquid partly or wholly it experiences an upward force which is called buoyant force.
The amount of buoyant force depends on the volume of body immersed, density of liquid and the value of acceleration due to gravity.
Here, the density of liquid is same in both the cases and g be the same. So, here the amount of buoyant force depends on the volume of body immersed.
As the density of lead is more than the density of aluminium, so the volume of aluminium is more than lead, as volume is equal to mass divided by density. So, the buoyant force acting on the aluminium is more than lead.
Answer:
120,000
Explanation:
Millimeters to meters calculation-
Multiply by 1,000.
120 x 1,000 = 120,000.
This is the correct answer and formula.
Hope this helps!
Answer:
B: False
Explanation:
The second law of thermodynamics states that: the entropy of an isolated system will never decrease because isolated systems always tend to evolve towards thermodynamic equilibrium which is a state with maximum entropy.
Thus, it means that the entropy change will always be positive.
Therefore, the given statement in the question is false.
Answer:

Explanation:
The formula for the force exerted between two charges is

where k is the Coulomb constant.
The charges are identical, so we can write the formula as

