Final answer:
To raise the temperature of 125.0 g of lead from 17.5°C to 41.1°C, 1250.0 calories of heat is needed.
Step-by-step explanation:
To calculate the heat required to raise the temperature of lead, we can use the formula:
Q = mc∆T
Where:
- Q is the heat in calories
- m is the mass of the lead in grams (125.0 g)
- c is the specific heat of lead (0.130 J/g°C)
- ∆T is the change in temperature (41.1°C - 17.5°C)
Plugging in the values, we get:
Q = (125.0 g)(0.130 J/g°C)(41.1°C - 17.5°C)
Q = 1250.0 calories
Therefore, the heat needed to raise the temperature of 125.0 g of lead from 17.5°C to 41.1°C is 1250.0 calories.