In a certain process a block of lead with a mass of 150g undergoes a temperature drop of 55.0 Celsius. In the process of 1072 J of heat energy is released from the Lead. What is the specific heat of Lead.
To find the specific heat of lead, you can use the formula:
\(q = mcΔT\)
Where:
- q is the heat energy released (1072 J)
- m is the mass of the lead (150 g or 0.150 kg)
- c is the specific heat capacity of lead (what we want to find)
- ΔT is the temperature change (-55.0 °C)
Rearranging the formula, we have:
\(c = \frac{q}{mΔT}\)
Substituting the given values:
\(c = \frac{1072 \, \text{J}}{0.150 \, \text{kg} \times \(-55.0 \, \text{°C}\)}\)
First, convert the mass from grams to kilograms:
\(m = 150 \, \text{g} = 0.150 \, \text{kg}\)
Now, substitute the values:
\(c = \frac{1072 \, \text{J}}{0.150 \, \text{kg} \times \(-55.0 \, \text{°C}\)}\)
Calculating the specific heat:
\(c = \frac{1072 \, \text{J}}{\(-8.25 \, \text{°C} \times 0.150 \, \text{kg}\)}\)
Finally, solve for c:
\(c = \frac{1072 \, \text{J}}{\(-8.25 \times 0.150\)}\)