- #1

- 6

- 0

## Homework Statement

A 155-g piece of copper at 188oC is dropped into 250.0 g of water at 23.7oC. (The specific heat of copper is 0.385 J/goC.) Calculate the final temperature of the mixture. (Assume no heat loss to the surroundings.)

**Knowns.**

155g Copper Initial Temp 188 C

250.0g Water Initial Temp 23.7 C

Spec Heat Copper .385 J/G C

Spec Heat Water 4.184 J / G C

**Unknown**

Final Temp Water

C - Celsius

J - Joules

G - Grams

T - Final Temp

## Homework Equations

Heat lost by copper = Heat gained by Water

my equation setup

155g × 0.385J / g °C × (Tfinal-188 °C) = 250.0g × 4.184j / g °C × (Tfinal-23.7 °C)

## The Attempt at a Solution

Well I have several pages of attempts here but I'll take a screenshot of my most recent.

Oddly enough I took a guess on the system we use (wileyplus) to get a hint as it gives you a hint on attempt 2 and got it correct. But it's bugging me that I can't solve it even after repeated attempts.

imageshack.us/photo/my-images/38/hwquestion.jpg/

^ wiley Plus screen saying my 32.5 guess was correct /

**This attempt I dropped the units and just solved for T**

Someone in the class mentioned trying this method but the result is off or I'm completely brain dead this evening and failing at math /

155g x 0.385 J/g C x (Tfinal-188 C) = 250.0g x 4.184 J/g C x (Tfinal-23.7 C)

59.675 x (t - 188) = 1046 x (t - 23.7)

59.675t - 11218.9 = 1046t - 24790.2

-986.325t = -13571.3

t = 13.759

Which would be way to far off from the +-1 sigfig on the 3rd digit allowance that said 32.5 C was correct.

When I go for the dimensional analysis route I can't isolate Celsius, I always end up with joules as a unit on my final answer.

155g | .385 J | (T - 188 C) | = 250g | 4.184 J | (T - 23.7 C)

| g C | | | g C |

Gives me the same result with Joules left.

I'm totally lost here, the bad part of my brain keeps saying you guessed the right answer so move on, but I can't leave this alone I have to know how to solve it. Any help would be much appreciated.

Last edited: