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?

You are watching: What is the final temperature of the water 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