A 110 kg sample of water at 130c is in a calorimeter you...

A 1.10 kg sample of water at 13.0°C is in a calorimeter. You drop a piece of steel with a mass of 0.390 kg at 240°C into it. After the sizzling subsides, what is the final equilibrium temperature (in °C)? (Make the reasonable assumptions that any steam produced condenses into liquid water during the process of equilibration and that the evaporation and condensation don't affect the outcome.)

Transcript

00:01 Alrighty, so we have a sample of water.

00:04 I'm just going to say water here.

00:07 And this sample as a mass m slash w equal to 1 .28 kilograms.

00:17 All right, and we know it's at 10 degrees celsius.

00:23 All right.

00:25 And we know that we are dropping a piece of steel into this water.

00:33 So steel.

00:35 Which naturally goes into the water.

00:38 But nonetheless, we have m sub s for m sub steel, and that is 0 .385 kilograms.

00:47 And this guy is at 215 degrees celsius.

00:52 All right.

00:54 So after all the sizzling subsides, there is an equilibrium temperature in this situation, and that's what we're curious about here.

01:01 We are curious on the t final.

01:05 All right and that is something that is going to be unique or is going to be the same for both the steel and the water since they're going to be the equilibrium they're both going to share the same final temperature what differs regarding temperature is their initial temperatures which affects their delta t but in regards to their final that's the same so that's something we want to keep in the back of our mind there so let's go ahead and figure out that final temperature so we're remember, very important equation here in thermodynamics.

01:42 My heat, q, is equal to mass times the specific heat capacity, times the change in temperature, which we could write out as mass times c times t final minus t initial.

01:55 All right.

01:56 Now, with our situation here, we actually have two of these equations.

02:01 We have one for water, q sub w, and that'll equal the mass of the water, multiplied by the specific heat of the water, and then t2 or t -final minus t -initial.

02:17 All right, and this is t -initial for the water.

02:20 So i'm going to be specific and label that.

02:22 The t -finals are going to be the same for both substances, so i don't have to bother writing water there.

02:27 But we will have t -initial, and then a sub -little there, so we know it's for the water.

02:34 And then the same for the steel, just different values here.

02:40 T final for the steel, which is the same as t final for the water, minus, t initial for the steel.

02:47 All right, so we have two equations there...