Question

In a coffee-cup calorimeter, 100.0 g of H2O and 100.0 mL of HCl are mixed. The HCl had an initial temperature of 44.6 C and the water was originally at 24.6 C. The reaction reaches thermal equilibrium (Ie. both substances have the same temperature) after 2800J is absorbed by water. What is the final temperature of water? (tenth)What is the specific heat of HCl? J/gC (tenth)

Answer 1

Given:

Mass of H20 = 100.0 g

Volume of HCL = 100.0 mL

Initial temperature of HCL = 44.6° C

Initial temperature of water (H20) = 24.6° C

The reaction reaches thermal equilibrium after 2800J is absorbed by water(H20).

Let's solve for the following:

• (a). What is the final temperature of water?

Since the reaction reaches thermal equilibrium, this means the final temperature of both substances will be the same.

To find the final temperature, apply the formula:

`Q=mc\Delta T`

Where:

Q is the heat absorbed by water = 2800 J = 2.8 kJ

m is the mass of water in kg = 100.0 g = 0.100 kg

ΔT is the temperature change.

c is the specific heat of water = 4184 J/kg C

Now, let's solve for ΔT:

`\begin{gathered} 2800=0.1*4184*\Delta T \\ \\ \Delta T=(2800)/(0.1*4184) \\ \\ \Delta T=6.69^0C \end{gathered}`

The temperature change is 6.69° C.

Now, to find the final temperature of water, we have:

Final temperature of water = Initial temperature of water + temperature change

Final temperature = 24.6° C + 6.69° C = 31.29° C = 31.3° C

Therefore, the final temperature of water is 31.3° C

• (b). What is the specific heat of HCl?

To find the specific heat of HCl, still apply the formula:

`Q=mc\Delta T`

We know that:

Heat lost = Heat gained

Where:

c is the specific heat of HCL

M(HCL) is the mass of HCL = 100.0 mL = 100g

Q = 2800 J

ΔT = 44.6° C - 6.69° C = 37.91° C

Now, for heat lost by HCL:

`100*c*37.91=3791*c`

For heat gained by H20 = 2800 J

Thus, to find the specific heat of HCL, we have:

`c=(2800)/(3791)=0.7\text{ J/g }^0C`

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