Question

The data in the table below were obtained during the Cheeto lab, in which a Cheeto was burned under a can of water, and the temperature change of the water was observed. What was the heat stored in the Cheeto in Cal? Round your answer to the nearest 0.01.1 Calorie = 1,000 calories4.184 J = 1 calSpecific heat of water is 4.184 J/g °C.They collect the following data:(the data is the image attached)

Answer 1

0.16 Cal.

Step-by-step explanation:

What is given?

Specific heat of water (c) = 4.184 J/g °C.

Mass of water in can (m) = 48.6 g.

Change of temperature (ΔT) = 24.5 °C - 21.3 °C = 3.2 °C.

What do we need? The heat storedin Cheeto (Q).

Ste-by-step siolution:

To solve this problem, e have to use the following formula:

`Q=c\cdot m\cdot\Delta T.`

Where Q is heat, c is the specific heat, m is the mass and ΔT is the change of temperature.

We just have to replace the given values in the formula:

`\begin{gathered} Q=4.184\text{ }\frac{J}{g\text{ }\cdot\degree C}\cdot48.6\text{ g}\cdot3.2\degree C, \\ Q=650.7\text{ J.} \end{gathered}`

But we have to give the answer in units of Cal. Remember that 4.184 J equals 1 cal:

`650.70\text{ J}\cdot\frac{1\text{ cal}}{4.184\text{ J}}=155.52\text{ cal.}`

But we want to find Cal. Remember that 1 Calorie equals 1000 calories. The conversion will look like this:

`155.2\text{ cal}\cdot\frac{1\text{ Calorie}}{1000\text{ cal}}=0.155\text{ Calorie \lparen Cal\rparen.}`

The answer is that te heat stored in the Cheeto is 0.156Cal.