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Carrie is trying to figure out the number of calories in a cube of cheese. To do this, she pours 154.7 mL of water into an aluminum can suspended from a ring stand. She takes the temperature of the water, and finds it to be 16.3 degrees Celsius. Then, she places the 5.23 gram cube of cheese under the can and lights it on fire! While the cheese is burning and for a few minutes after it is done, Carrie records the temperature of the water, finding that it levels out at 24.1 degrees Celsius. How many calories of heat were gained by the water? Please answer to the nearest 0.1 calorie.

Sagot :

To calculate the calories that the water gained, we must use the specific heat of the water. The specific heat of water is 1cal/g°C, which means that it takes one calorie to raise one gram of water 1°C.

Now, we need the mass of water and the temperature difference.

We will calculate the mass of water from its density, the density of water is 1mL/g. Density is defined as:

[tex]Density=\frac{Mass}{Volume}[/tex]

We clear the mass and replace the known data:

[tex]Mass=1g/ml\times154.7mL=154.7g[/tex]

Now, the temperature difference will be:

[tex]\begin{gathered} \Delta T=T_2-T_1 \\ \Delta T=24.1\degree C-16.3\degree C=7.8\degree C \end{gathered}[/tex]

Calories gained by water are calculated from the following equation:

[tex]Q=mCp\Delta T[/tex]

where,

Q is the heat or energy that the substance absorbs or releases. In this case, the water will absorb energy.

m is the mass of water = 154.7g

Cp is the specific heat of water, 1cal/g°C

dT is the difference of temperature, 7.8°C

Now, we replace the known data:

[tex]\begin{gathered} Q=154.7g\times1\frac{cal}{g.\degree C}\times7.8\degree C \\ Q=1206.7cal \end{gathered}[/tex]

Answer: The water gained 1206.7 calories of heat

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