3. Calculate the specific heat of water if we need 4200 J to change the temperature of 1kg by 1 °C.

4. How much heat is necessary to heat up 0.05 kg of Cobalt from 30 °C to 40°C? The specific heat of Cobalt is 0.41868 (°C)

### Problem 3: Specific Heat of Water

The specific heat capacity (\(c\)) of a substance is given by the formula:

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

Where:
- \(Q\) is the heat energy added (in joules, J),
- \(m\) is the mass of the substance (in kilograms, kg),
- \(\Delta T\) is the change in temperature (in degrees Celsius, °C).

Given:
- \(Q = 4200\) J,
- \(m = 1\) kg,
- \(\Delta T = 1\) °C.

Now, we can calculate the specific heat capacity of water:

\[
c = \frac{4200 \, \text{J}}{1 \, \text{kg} \times 1 \, \text{°C}} = 4200 \, \text{J/kg°C}
\]

So, the specific heat of water is **4200 J/kg°C**.

---

### Problem 4: Heat Required to Heat Cobalt

The heat energy required (\(Q\)) to change the temperature of a substance is given by:

\[
Q = m \cdot c \cdot \Delta T
\]

Where:
- \(m = 0.05\) kg (mass of Cobalt),
- \(c = 0.41868 \, \text{J/kg°C}\) (specific heat of Cobalt),
- \(\Delta T = 40°C - 30°C = 10°C\) (change in temperature).

Now, calculate the heat energy:

\[
Q = 0.05 \, \text{kg} \times 0.41868 \, \text{J/kg°C} \times 10 \, \text{°C} 
\]
\[
Q = 0.20934 \, \text{J/°C} \times 10 \, \text{°C} = 2.0934 \, \text{J}
\]

Thus, the heat required to heat up 0.05 kg of Cobalt from 30°C to 40°C is **2.0934 J**.

---

Do you want more details on these calculations or have any questions?

**Related Questions:**
1. What is the physical meaning of specific heat capacity?
2. How does specific heat capacity affect the rate at which a substance heats up?
3. Can specific heat capacity vary with temperature?
4. How much heat is required to heat up 1 kg of Cobalt by 1°C?
5. How do you experimentally determine the specific heat capacity of a substance?

**Tip:** Specific heat capacity is a material-specific property that indicates how much heat energy is needed to change the temperature of a unit mass by one degree. Substances with high specific heat capacities, like water, require more energy to change temperature.

Learn how to calculate the specific heat of water and the heat energy required to raise the temperature of Cobalt. Specific examples include using the specific heat capacity formula and applying it to solve practical temperature change problems. Suitable for high school level students.

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