A 500.0 g iron bar at 212 C is placed in 2.0L of water at 24.0 C. What will be the change in temperature water

Question

Chemistry

iron bar 0.473j/gC water 4.184j/gC

Answer 1

A couple of assumptions are necessary:

1) density of water is 1 g/mL.
2) All the heat released by the iron bar goes into heating the water.

From assumption 1: 2.0L of water = 2000 mL of water (since 1L=1000mL). Thus, the mass of water is: 2000mL x 1g/mL = 2000g

Assumption 2 can be expressed as:

Heat released from Iron + Heat absorbed by water = 0 (1)

Q=m * Specific heat * Change in Temperature (where Q is heat) (2)

Where change in temperature= Final T - Initial T. (3)

Let's call the final temperature of the system (both iron and water) T. Then we have:

I am going to omit units so the equation is more clear. This should be done only when you are sure that units are consistent among all the quantities, which is the case in this problem.

Using equation (2) in equation (1) we have:

500 * 0.473 * (T-212) + 2000 * 4.184 * (T-24) = 0

8368T -200832+236.5T - 50138 = 0

solving for T we find: T=29.2 C

so, the system reaches equilibrium at 29.2 C. The change in the water temperature is then:

29.2-24=5.2 C

and for iron is 212-29.2=182.8 C

Note the huge difference in change in T between the two parts of the system. Water has a relatively large heat capacity and it is able to absorb large amounts of heat (or release them) with relatively little changes in temperature.

Answer 2

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Answer 3

Tell me the heat capacity of your iron bar and the heat capacity of water and I will give you the answer.