Question

Calculate the specific heat capacity for a 22.7-g sample of lead that absorbs 237 J when its temperature increases from 29.8 °C to 95.6 °C.

Answer 1

Answer:

$\boxed {\boxed {\sf c\approx 0.159 \ J/ g \textdegree C}}$

Explanation:

We are asked to find the specific heat capacity of a sample of lead. The formula for calculating the specific heat capacity is:

$c= \frac{Q}{m \times \Delta T}$

The heat absorbed (Q) is 237 Joules. The mass of the lead sample (m) is 22.7 grams. The change in temperature (ΔT) is the difference between the final temperature and the initial temperature. The temperature increases from 29.8 °C to 95.6 °C.

• ΔT = final temperature -inital temperature
• ΔT= 95.6 °C - 29.8 °C = 65.8 °C

Now we know all three variables and can substitute them into the formula.

• Q= 237 J
• m= 22.7 g
• ΔT = 65.8 °C

$c= \frac {237 \ J}{22.7 \ g \ \times \ 65.8 \textdegree C}$

Solve the denominator.

• 22.7 g * 65.8 °C = 1493.66 g °C

$c= \frac {237 \ J}{1493.66 \ g \textdegree C}$

Divide.

$c= 0.1586706479 J /g \textdegree C$

The original values of heat, temperature, and mass all have 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place tells us to round the 8 up to a 9.

$c \approx 0.159 \ J/g \textdegree C$

The specific heat capacity of lead is approximately 0.159 Joules per gram degree Celsius.