Question

a chemist carefully measures the amount of heat needed to raise the temperature of a 437.0 g sample of a pure substance from -5.7°c to 9.0°c. the experiment shows that 893. j of heat are needed. what can the chemist report for the specific heat capacity of the substance? be sure your answer has the correct number of significant digits.

Explanation:

Step1: Recall the formula for heat

The formula for heat \( q \) is \( q = mc\Delta T \), where \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is the change in temperature. We need to find \( c \), so rearrange the formula: \( c=\frac{q}{m\Delta T} \).

Step2: Calculate the change in temperature

The temperature change in Celsius is the same as in Kelvin (since the size of a Celsius degree is the same as a Kelvin degree). So \( \Delta T = T_2 - T_1 = 9.0^{\circ}\text{C}-(-5.7^{\circ}\text{C}) = 14.7^{\circ}\text{C}=14.7\text{ K} \).

Step3: Plug in the values

We know \( q = 893\text{ J} \), \( m = 437.0\text{ g} \), and \( \Delta T = 14.7\text{ K} \). Substitute into the formula for \( c \):
\( c=\frac{893\text{ J}}{437.0\text{ g} \times 14.7\text{ K}} \)

Step4: Perform the calculation

First, calculate the denominator: \( 437.0\times14.7 = 437.0\times14 + 437.0\times0.7 = 6118 + 305.9 = 6423.9 \)
Then, \( c=\frac{893}{6423.9}\approx0.139\text{ J g}^{-1}\text{ K}^{-1} \) (considering significant digits: \( q \) has 3 significant digits, \( m \) has 4, \( \Delta T \) has 3, so the result should have 3 significant digits).

Answer:

\( 0.139 \)