The same energy Q enters five different substances as heat. Which of these has the greatest specific heat ?

Options

A. The temperature of 3g of substance A increases by 10 K.
B. The temperature of 4g of substance B increases by 4K.
C. The temperature of 6g of substance C increases by 15K.
D. The temperature of 8g of substance D increases by 5K.

Answer

To determine which of the five substances has the greatest specific heat when the same amount of energy \( Q \) enters each substance, we can use the formula for heat transfer:

\[ Q = m \cdot s \cdot \Delta \theta \]

Where:
- \( Q \) = heat energy (in joules)
- \( m \) = mass of the substance (in grams)
- \( s \) = specific heat capacity (in J/g·K)
- \( \Delta \theta \) = change in temperature (in Kelvin)

### Step-by-Step Solution:

1. **Identify the Formula**:
We start with the formula for heat transfer:
\[ Q = m \cdot s \cdot \Delta \theta \]

2. **Rearranging the Formula**:
To find the specific heat \( s \), we rearrange the formula:
\[ s = \frac{Q}{m \cdot \Delta \theta} \]

3. **Calculate Specific Heat for Each Substance**:
- **Substance A**:
- Mass \( m_A = 3 \, \text{g} \)
- Change in temperature \( \Delta \theta_A = 10 \, \text{K} \)
- Specific heat \( s_A = \frac{Q}{3 \cdot 10} = \frac{Q}{30} \)

- **Substance B**:
- Mass \( m_B = 4 \, \text{g} \)
- Change in temperature \( \Delta \theta_B = 4 \, \text{K} \)
- Specific heat \( s_B = \frac{Q}{4 \cdot 4} = \frac{Q}{16} \)

- **Substance C**:
- Mass \( m_C = 9 \, \text{g} \)
- Change in temperature \( \Delta \theta_C = 10 \, \text{K} \)
- Specific heat \( s_C = \frac{Q}{9 \cdot 10} = \frac{Q}{90} \)

- **Substance D**:
- Mass \( m_D = 4 \, \text{g} \)
- Change in temperature \( \Delta \theta_D = 10 \, \text{K} \)
- Specific heat \( s_D = \frac{Q}{4 \cdot 10} = \frac{Q}{40} \)

- **Substance E**:
- Mass \( m_E = 5 \, \text{g} \)
- Change in temperature \( \Delta \theta_E = 5 \, \text{K} \)
- Specific heat \( s_E = \frac{Q}{5 \cdot 5} = \frac{Q}{25} \)

4. **Compare Specific Heats**:
Now we compare the specific heats calculated:
- \( s_A = \frac{Q}{30} \)
- \( s_B = \frac{Q}{16} \)
- \( s_C = \frac{Q}{90} \)
- \( s_D = \frac{Q}{40} \)
- \( s_E = \frac{Q}{25} \)

To find the greatest specific heat, we look for the smallest denominator:
- \( 30, 16, 90, 40, 25 \)

The smallest denominator is \( 16 \) (for substance B).

5. **Conclusion**:
Therefore, the substance with the greatest specific heat is **Substance B**.