Question

The masses and specific heats of some samples of liquids are shown in the table below.

Samples Mass (kg) Specific Heat Capacity (J/kg x K)
Water 0.75 4,200
Glycerin 0.75 2,400
Methanol 0.75 2,500
Cooking Oil 0.75 2,100

The temperature of which sample will raise most when 1,000J of heat is added?
Responses
A MethanolMethanol
B GlycerinGlycerin
C Cooking oilCooking oil
D Water

Answer 1

To determine which sample will have the highest temperature change when 1,000 J of heat is added, you can use the formula:

\[
\Delta T = \frac{Q}{m \cdot c}
\]

where:
- \(\Delta T\) is the change in temperature (in K),
- \(Q\) is the heat added (in J),
- \(m\) is the mass of the substance (in kg),
- \(c\) is the specific heat capacity (in J/kg·K).

Let's calculate the temperature change for each sample.

1. **Water**
- \(m = 0.75\) kg
- \(c = 4,200\) J/kg·K
\[
\Delta T_{\text{water}} = \frac{1000}{0.75 \times 4200} = \frac{1000}{3150} \approx 0.317 \text{ K}
\]

2. **Glycerin**
- \(m = 0.75\) kg
- \(c = 2,400\) J/kg·K
\[
\Delta T_{\text{glycerin}} = \frac{1000}{0.75 \times 2400} = \frac{1000}{1800} \approx 0.556 \text{ K}
\]

3. **Methanol**
- \(m = 0.75\) kg
- \(c = 2,500\) J/kg·K
\[
\Delta T_{\text{methanol}} = \frac{1000}{0.75 \times 2500} = \frac{1000}{1875} \approx 0.533 \text{ K}
\]

4. **Cooking Oil**
- \(m = 0.75\) kg
- \(c = 2,100\) J/kg·K
\[
\Delta T_{\text{cooking oil}} = \frac{1000}{0.75 \times 2100} = \frac{1000}{1575} \approx 0.634 \text{ K}
\]

Now, let's summarize the results:

- \(\Delta T_{\text{water}} \approx 0.317 \text{ K}\)
- \(\Delta T_{\text{glycerin}} \approx 0.556 \text{ K}\)
- \(\Delta T_{\text{methanol}} \approx 0.533 \text{ K}\)
- \(\Delta T_{\text{cooking oil}} \approx 0.634 \text{ K}\)

The sample that will raise the most in temperature when 1,000 J of heat is added is **Cooking Oil**.

**Answer: C. Cooking Oil**