Here is my work:eginalignQ &= mc,Delta T\Q_1 &= (pu30 g)(pu0.449 J g-1 ^circ C-1)(x - pu144 ^circ C) agIron\Q_2 &= (pu40 g)(pu4.184 J g-1 ^circ C-1)(x - pu20 ^circ C) agWater\ extSince, quad Q_1 &= -Q_2\13.47 (x-144) &= - (167.36)(x-20) puJ\13.47x - 1939.68 &= -167.36x + 3347.20\180.83x &= pu5286.88 ^circ C \x &= pu0.03420 ^circ Cendalign
This is providing me response that isn"t exactly according to my book. What go I execute wrong, and also how deserve to I solve it?
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All girlfriend did is essentially right, your only mistake is in the critical step, together LDC3 currently pointed the end in the comments. However, i am encouraging you to use systems all the method and when taking care of thermodynamics usage Kelvin rather of Celsius.eginalignQ &= mcDelta T\endalignNow friend can type the equations for each the the problem, if substituting $Delta T$ v a temperature range, gift $x$ the final temperature the whole system will finish up on.
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Additionally note, the the iron will certainly be cooled down, if the water will certainly be heated. (I am using a different strategy than you.eginalignQ_mathrmloss &= m(ceFe)cdotc(ceFe)cdot
The moved heat needs to equal $$Q_mathrmgain = Q_mathrmloss$$
With this you can solve for $x$.eginalignm(ceFe)cdotc(ceFe)cdot