[whitley881]

A volume of 90.0 of is initially at room temperature (22.0). A chilled steel rod at 2.00 is placed in the water. If the final temperature of the system is 21.2, what is the mass of the steel bar?
Use the following values:

specific heat of water = 4.18

specific heat of steel = 0.452

[jcaron2]

You really should specify the units when you right down your numbers. I'm assuming, for example, that your 90 volume units of water is in cm^3 (as opposed to quarts, gallons, cubic meters, etc.). Also, I assume your temperature must be in Celsius.

Okay, so what's the mass of 90 cm^3 of water? Well, the density of water is 1g/cm^3, so the mass is



So you have 90 grams of water (which makes sense since a gram is, by its very definition, the mass of 1 cm^3 of water).

The water was chilled by 0.8°C (it started at 22°C and ended at 21.2°C, for a net change of -0.8°C). It got cooler because it transferred some of its heat to the steel bar. The water cooled down and the steel bar warmed up until their temperatures became equal.

So how much energy did the water give to the steel? We know the mass of the water, the change in temperature, and the specific heat, so it's straightforward to calculate the energy:



I'll let you do the math to figure out the exact number of joules of energy the water gave up.

Once you know how much energy the water lost, you automatically know how much energy the steel bar gained. It's the same number. So now you just have to repeat the same formula as above, except this time you know everything except m. The energy, Q, is the same as what the water gave up. The specific heat, c, is 0.452 J/g°C. And the change in temperature is 21.2°C - 2°C = 19.2°C. All you have to do is solve for m. Can you do that?