Two liquids, A and B, have equal masses and equal initial temperatures. Each is heated for the same length of time over identica
1 answer:
Answer:
So the specific heat of the liquid B is greater than that of A.
Explanation:
Liquid A is hotter than the liquid B after both the liquids are heated identically for the same duration of time from the same initial temperature then according to heat equation,
where:
m = mass of the body
c = specific heat of the body
change in temperature of the body
The identical heat source supplies the heat for the same amount of time then the quantity of heat supplied is also equal.
So for constant heat, constant mass the temperature change is inversely proportional to the specific of heat of the liquid.
So the specific heat of the liquid B is greater than that of A.
You might be interested in
A. 0.77 A
Using the relationship:
where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.
For the first light bulb, P = 60 W and V = 120 V, so the resistance is
For the second light bulb, P = 200 W and V = 120 V, so the resistance is
The two light bulbs are connected in series, so their equivalent resistance is
The two light bulbs are connected to a voltage of
V = 240 V
So we can find the current through the two bulbs by using Ohm's law:
B. 142.3 W
The power dissipated in the first bulb is given by:
where
I = 0.77 A is the current
is the resistance of the bulb
Substituting numbers, we get
C. 42.7 W
The power dissipated in the second bulb is given by:
where
I = 0.77 A is the current
is the resistance of the bulb
Substituting numbers, we get
D. The 60-W bulb burns out very quickly
The power dissipated by the resistance of each light bulb is equal to:
where
E is the amount of energy dissipated
t is the time interval
From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.