What is the final temperature the water and also iron if a $pu30 g$ item of iron in ~ $pu144 °C$ to be dropped into a calorimeter through $pu40 g$ the water at $pu20 °C$?Specific heat of water is $pu4.184 J g-1 ^circ C-1$, and also of iron is $pu0.449 J g-1 ^circ C-1$

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\Q_mathrmgain &= m(ceH2O)cdotc(ceH2O)cdot\endalign

The moved heat needs to equal $$Q_mathrmgain = Q_mathrmloss$$

With this you can solve for $x$.eginalignm(ceFe)cdotc(ceFe)cdot &= m(ceH2O)cdotc(ceH2O)cdot\%m(ceFe)cdotc(ceFe)cdotT(ceFe)-m(ceFe)cdotc(ceFe)cdotx &= m(ceH2O)cdotc(ceH2O)cdotx-m(ceH2O)cdotc(ceH2O)cdotT(ceH2O)\% m(ceFe)cdotc(ceFe)cdotT(ceFe) +m(ceH2O)cdotc(ceH2O)cdotT(ceH2O)&= m(ceH2O)cdotc(ceH2O)cdotx+m(ceFe)cdotc(ceFe)cdotx\% m(ceFe)cdotc(ceFe)cdotT(ceFe) +m(ceH2O)cdotc(ceH2O)cdotT(ceH2O)&= cdotx\x &=fracm(ceFe)cdotc(ceFe)cdotT(ceFe) +m(ceH2O)cdotc(ceH2O)cdotT(ceH2O) m(ceH2O)cdotc(ceH2O)+m(ceFe)cdotc(ceFe)\%x &=frac30~mathrmgcdot0.449~mathrmfracJgKcdot417~mathrmK +40~mathrmgcdot4.184~mathrmfracJgKcdot293~mathrmK 40~mathrmgcdot4.184~mathrmfracJgK+30~mathrmgcdot0.449~mathrmfracJgK\x &= frac5616.99~mathrmJ+49036.48~mathrmJ167.36~mathrmfracJK+13.47~mathrmfracJK\x &= frac54653.47180.83~mathrmK =302.24~mathrmK\x &approx 29~mathrm^circCendalign