Steel blocks A and B, which have equal masses, are at TA = 300 oC and T8 = 400 oC. Block C, with mc - 2mA, is at TC = 350 oC. Bl

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Steel blocks A and B, which have equal masses, are at TA = 300 oC and T8 = 400 oC. Block C, with mc - 2mA, is at TC = 350 oC. Bl

ocks A and B are placed in contact, isolated, and allowed to come into equilibrium. Then they are placed in contact with block C. At that instant:a.TA = TB < TCb.TA = TB = TCc.TA = TB > TCd.TA + TB = TCe.TA - TB = TC

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1 answer:

serg [3.5K] · Answer 1 · 2026-02-27T20:57:47+03:00

Answer:

b) TA = TB = TC

Explanation:

When the blocks are brought into contact and isolated from the environment, they will exchange heat until they achieve thermal equilibrium.
During this exchange, the hotter body will lose heat, which will be gained by the cooler body.
The equilibrium state will be established once this equation is satisfied:

$\Delta Q = c_{st}* m_{A} * (T_{fin} - T_{0A} ) = c_{st}* m_{B} * (T_{0B} - T_{fin} )$

Substituting the initial temperatures T₀A = 300º C and T₀B = 400ºC, while simplifying for equal block masses mA = mB, enables us to solve for the final temperature, Tfin:

$(400 \ºC - T_{fin}) = (T_{fin} - 300 \ºC) \\ \\ 2* T_{fin} = 700\ºC\\ \\ T_{fin} = 350\ºC$

At equilibrium, when both blocks combine, they will yield a uniform final temperature of 350ºC.
When block C, also at this temperature, makes contact, all three blocks will simultaneously reflect this final temperature of 350 ºC.
Therefore, option b) is correct.