The specific heat of lead is 0.0380 cal/g·°C. If 47.0 calories of energy raised the temperature of a lead sample from 28.3°C to 30.1°C what is the mass of the sample? A) 26 g B) 687 g C) 1.2 × 10³ g D) 2.3 × 10³ g

Question

Quizplus · Accepted Answer

We can use the formula:
Q = mcΔT
Where Q is the heat energy transferred, m is the mass of the sample, c is the specific heat, and ΔT is the change in temperature.
Substituting the values, we get:
47.0 cal = m (0.0380 cal/g°C) (30.1°C - 28.3°C)
47.0 cal = m (0.0380 cal/g°C) (1.8°C)
47.0 cal = 0.0684 m cal
m = 47.0 cal / 0.0684 cal/g = 687 g
Therefore, the mass of the lead sample is 687 g, which is option B.

The Specific Heat of Lead Is 0