In an insulated vessel, a quantity of hot water at temperature T1 is mixed with a different quantity of cold water at temperatur
1 answer:
Answer:Water Only
Explanation:
Given
vessel is insulated therefore no heat can be added or removed i.e. heat exchange is zero
If hot water at is mixed with cold water at
then at equilibrium vessel contains only water and final temperature of water will be between
and
Heat released by hot water is equal to heat gain by cold water .
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<h2>Answer:</h2>
(a) 3.18Ω
(b) 3.18Ω
<h2>Explanation:</h2>Let the two bulbs be A and B
Given;
= Resistance in bulb A = 3.0Ω
= Resistance in bulb B = 2.5Ω
Since the two bulbs are connected in parallel;
i. their effective resistance () is given by
=
+
---------------(i)
Substitute the values of and
into equation (i)
=> =
+
Solve for
= 1.36Ω
ii. voltage (potential difference), V, across them is the same;
Therefore we can get the total current (I) that will flow through them if the voltage to be supplied is 2.4V.
Use the Ohm's law;
V = I x R -----------------(ii)
Where;
V = voltage across them = 2.4V
I = total current flowing through them
R = their effective resistance = = 1.36Ω
Substitute these values into equation (ii) as follows;
2.4 = I x 1.36
I = 2.4 / 1.36
I = 1.76A
(a) Now get the value of R
Since the voltage across the two bulbs is 2.4V out of the 8.0V supplied by the source, then the remaining (8.0 - 2.4 = 5.6)V will pass across the resistor R.
Also, since the two bulbs make a series connection with the resistor R, the same total current (I = 1.76A) that flows through these bulbs will flow through the resistor R.
Therefore, to get the value of R, we use the relation
V = I x R ------------------------------(iii)
Where;
V = voltage across the resistor = 5.6V
I = current through the resistor = 1.76A
<em>Substitute these values into equation (iii)</em>
=> 5.6 = 1.76 x R
=> R = 5.6 / 1.76
=> R = 3.18Ω
Therefore, the value of R to be chosen in order to supply each bulb with a voltage of 2.4V is 3.18Ω
(b) The potential difference and voltage across refer to the same thing. Therefore, the value of R that would make the potential difference across each of the bulbs be 2.4V is the same as the one calculated in (a) which is 3.18Ω