Body A has twice the mass and three times the specific heat of body B. They are supplied with equal amounts of heat. Body A experiences a temperature change ΔT. What change in temperature is experienced by body B?
Question 2 options:
A) ΔT
B) 3ΔT/2
C) 2ΔT/3
D) 6ΔT
E) ΔT/2
heat=mass*specht*deltaT
deltatempB=heat/massb*specificheatB
deltatempB=6heat/massA*specifheatA=6deltTa
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6
To find the change in temperature experienced by body B, we can use the formula for heat transfer:
Q = mcΔT
where Q is the heat transferred, m is the mass, c is the specific heat, and ΔT is the change in temperature.
Let's assume that the heat transferred to both bodies A and B is the same. Let's also assume that the initial and final temperatures of both bodies are the same.
For body A, we can write:
Q = mA * cA * ΔTA
For body B, we can write:
Q = mB * cB * ΔTB
Since both bodies are supplied with equal amounts of heat, we have:
mA * cA * ΔTA = mB * cB * ΔTB
Given that mA = 2mB and cA = 3cB, we can substitute these values into the equation:
(2mB) * (3cB) * ΔTA = mB * cB * ΔTB
Simplifying, we can cancel out mB * cB from both sides of the equation:
6 * ΔTA = ΔTB
Therefore, the change in temperature experienced by body B is 6 times the change in temperature experienced by body A.
Thus, the correct answer is D) 6ΔT.