Question

calculate the heat capacity of 1,343 g of lead, given that 45 j is needed to raise the temperature by 29.8 ∘c. round your answer to the nearest tenth.

Answer 1

The heat capacity of 1,343 g of lead is 0.0 J/g·°C.

Step-by-step explanation:

The heat capacity of a substance is the amount of heat energy required to raise the temperature of that substance by one degree Celsius. To calculate the heat capacity (C) of lead, we can use the formula: C = Q / (m * ΔT), where Q is the amount of heat energy, m is the mass of the substance, and ΔT is the change in temperature.

In this case, the heat (Q) needed to raise the temperature of lead by 29.8 °C is given as 45 J. The mass of lead is 1,343 g. Plugging these values into the formula, we get C = 45 J / (1,343 g * 29.8 °C).

Calculating the above expression gives us a value of approximately 0.001268 J/g·°C. Rounding this to the nearest tenth, the heat capacity of 1,343 g of lead is 0.0 J/g·°C.

Answer 2

The heat capacity of 1,343 g of lead, with 45 J needed to raise the temperature by 29.8 °C, is approximately 1.5 J/°C when rounded to the nearest tenth.

Step-by-step explanation:

To calculate the heat capacity of 1,343 g of lead with the information that 45 J is needed to raise its temperature by 29.8 °C, we use the formula:

Heat Capacity (C) = Q / ΔT

Where:

• Q is the amount of heat absorbed or released (45 J in this case), and
• ΔT is the change in temperature (29.8 °C in this case).

Plugging in the values:

C = 45 J / 29.8 °C
C = 1.51006711409396 J/°C

Round the answer to the nearest tenth:

C ≈ 1.5 J/°C