Question

A 78.0 sample of gold at 175°C is put in a foam cup calorimeter with 25.0 g of water at 25.0°C. Calculate the specific heat of the gold if the final temperature is 38.18°C and the specific heat of water is 4.18 J/g°C. Show your work.
Thanks for the help!!​

Answer 1

The answer is the specific heat of the gold is 0.129 J/gC .

What is Calorimetry ?

It is an experimental method that allows one to calculate the heat change in a chemical process.

Calorimeter is just a reaction vessel. It could be a foam cup, a soda can, or a commercially available bomb calorimeter

Where q represents the heat change in the reaction, the calorimeter and the water. Since you are only measuring temperature, you will need to calculate the heat using it.

The change in heat of the gold is given by:

$\rm q_{gold} = m c_{gold}\Delta T$

$=78 \times c_{p}\times (175-38.18)$

The change in heat of the water is given by:

$\rm q_{water} = mc_{p}\Delta T$

m is the mass of water in the calorimeter in grams,

and delta T is the change in temperature.

where $\rm c_{p}$ is the specific heat of water, which is 4.184 J/gC

mass of water =  25 g of water.

Temperature T₁ = 25°C

Final Temperature = 38.18 °C

Difference between temperature = 13.18 °C

$\rm q_{water} = 25 \times 4.18\times 13.18$

For a Foam Cup Calorimeter

$\rm q_{water} = q_{cal}\\\\\\\rm 25\times4.18\times13.318 = 78 \times c_{p}\times (175-38.18)\\\\\\On \;solving\; this \;we \;get \;c_{p} = 0.129$

Therefore the  the specific heat of the gold is 0.129 J/gC

To know more about Calorimeter

brainly.com/question/4802333

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