Mass of iron 46.70 g Initial temperature of...

Mass of iron 46.70 g Initial temperature of Fe 79.10°C Volume of water 54.73 mL Initial temperature of water 20.30°C Final water temperature 24.90°C Density of water 1.00 $\frac{g}{mL}$ Specific heat of water 4.184 $\frac{J}{g \cdot °C}$ Specific heat of iron 0.444 $\frac{J}{g \cdot °C}$ A student transfers a heated sample of pure iron metal to a Styrofoam coffee cup calorimeter containing deionized water. Use the collected data to answer the following questions. (a) Assuming that heat was transferred from the iron to the water and the calorimeter, determine the calorimeter constant. Calorimeter constant = $\boxed{\frac{J}{°C}}$ (b) The actual calorimeter constant is 18.94 $\frac{J}{°C}$. Calculate the percent error for the experimentally determined calorimeter constant. Percent error = $\boxed{\%}$

Transcript

00:01 Hello, so here the question is based on calorimetry.

00:04 Okay, now in this question we are provided with the mass of iron as 44 .3 gram, then the initial temperature of iron is given as 79 .10 degree celsius, then we are provided with the volume of water as 54 .73 ml, which when converted into liters it will be 0 .05473 liters, then the initial temperature of water is given as 20 .30 degree celsius and the final temperature of water is given as 24 .60 degree celsius, then the density is given as 1 gram per milliliter, then the specific heat of water is given as 4 .184 joule per gram degree celsius, then we are also provided with the specific heat of iron as 0 .444 joule gram degree celsius, okay.

01:14 So what we need to calculate here is for the calorimeter constant, okay.

01:21 So let's see the answer for this question.

01:25 Now, we can have the equation as heat given by iron, this will be equal to the heat taken by water plus the heat taken by calorimeter, okay.

01:52 So heat given by iron have the equation m multiplied by c iron multiplied by the change in temperature, okay.

02:02 So this will be equal to heat taken up by water, so that will be mass multiplied by the heat capacity of water multiplied by the change in temperature, okay, plus the heat constant of calorimeter that is c multiplied by the change in temperature will give you the heat taken up by the calorimeter, okay.

02:24 So we can substitute all these values and calculate, so we have the mass of iron as 44 .3 grams, then this multiplied by 0 .444 and that is the heat capacity of iron multiplied by the change in temperature.

02:38 So the initial temperature is given as 79 .10 minus the final temperature of water will be the final temperature of the iron, so that is equal to 24 .60, okay.

02:53 So this is equal to the mass is given as 54 .73 because we are taking here the ml in terms of grams which means 1 ml is equal to 1 gram, okay.

03:06 Then 54 .73 ml will be equal to 54 .73 grams, okay...

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