The temperature of a 15-g sample of lead metal increases from 22 °C to 37 °C upon the addition of 29.0 J of heat.The specific heat capacity of the lead is ________ J/g-K.
A)7.8
B)1.9
C)29
D)0.13
E)-29

Question

Quizplus · Accepted Answer

The specific heat capacity (c) can be calculated using the formula $q = mc\Delta T$, where $q$ is the heat added, $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature. Rearranging for $c$, we get $c = \frac{q}{m\Delta T}$. Plugging in the given values, $c = \frac{29.0\, \text{J}}{15\, \text{g} \times (37\, \text{°C} - 22\, \text{°C})} = \frac{29.0}{15 \times 15} = 0.1293\, \text{J/g-K}$, which rounds to 0.13 J/g-K.

The Temperature of a 15-G Sample of Lead Metal Increases