Question

a sample of an unknown substance has a mass of 89.5 g. if 345.2 j of heat are required to heat the substance from 285 k to 305 k, what is the specific heat of the substance? use the formula ( q = mc_pdelta t ). (square) j/g·k

Explanation:

Step1: Identify known values

We know that \( q = 345.2 \, \text{J} \), \( m = 89.5 \, \text{g} \), \( T_1 = 285 \, \text{K} \), \( T_2 = 305 \, \text{K} \). First, calculate \( \Delta T \).
\( \Delta T=T_2 - T_1=305 - 285 = 20 \, \text{K} \)

Step2: Rearrange the formula for \( C_p \)

From \( q = mC_p\Delta T \), we can solve for \( C_p \) by dividing both sides by \( m\Delta T \):
\( C_p=\frac{q}{m\Delta T} \)

Step3: Substitute the values into the formula

Substitute \( q = 345.2 \, \text{J} \), \( m = 89.5 \, \text{g} \), and \( \Delta T = 20 \, \text{K} \) into the formula:
\( C_p=\frac{345.2}{89.5\times20} \)
First, calculate the denominator: \( 89.5\times20 = 1790 \)
Then, calculate the division: \( C_p=\frac{345.2}{1790}\approx0.193 \)

Answer:

\( 0.193 \)