- #1

Kaleb

- 49

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**[SOLVED] Quantity of Heat/Specific Heat Water+Iron**

## Homework Statement

Exact Problem:

What will be the final temperature of 100g of 20 degrees C water when 100g of 40 degrees C iron nails are submerged in it. (The specific heat of iron is .12 cal/g C. Here you should equate the heat gained by the water to the heat lost by the nails.)

## Homework Equations

Q = cm(change in)t

Q = quantity of heat

c = specific heat

(water = 4.184J/g or 1 cal/g)

(iron = .502J/g or .12 cal/g)

m = mass

100g + 100g = 200g total substance(water+iron)

t = temperature

(40(final)-20(initial))

## The Attempt at a Solution

I feel I slightly understand what I am supposed to do but the part where it says I need to find the heat lost by iron and the heat gained by water I am totally confused about. Heres my attempt: My assumption is that I need to do two different calculations and difference them out. Also if someone could point me in the right direction to be able to post real equations instead of what I have shown that would be so awesome.

Iron First:

Q = cm(Tf-Ti)

Q = .502J*100g(40-20)

Q = 1004J of Energy Lost

Water Second:

q = 4.184*100g(40-20)

Q = 8368J of Energy Gained

Answer:

8368J - 1004J = 7364J of Energy or 1760.04cal/g

I have tried numerous ways to find the temperature difference, lost and gained but I am clueless. Also I can't figure out where to transfer back into Celcius. By my calc's it would have gained like 1700 degrees haha. >.> I feel dumb... Anyways TIPS are always appreciated in advance. Constructive critism is appreciated as well, and well anything else that can make me a better poster is welcome too. ^^ thanks!