Question

the specific heat of water is 4.18 j/g·°c. how much heat does 225.0 g of water release when it cools from 85.5 °c to 50.0 °c? use the formula ( q = mc_pdelta t ). ( 2.90 \times 10^5 ) j ( 3.26 \times 10^4 ) j ( 1.27 \times 10^5 ) j ( 3.34 \times 10^4 ) j

Explanation:

Step1: Identify the values

We have \( m = 225.0\space g \), \( C_p = 4.18\space J/g\cdot^\circ C \), initial temperature \( T_i = 85.5^\circ C \), final temperature \( T_f = 50.0^\circ C \). First, calculate the change in temperature \( \Delta T = T_i - T_f \).
\( \Delta T = 85.5^\circ C - 50.0^\circ C = 35.5^\circ C \)

Step2: Apply the formula \( q = mC_p\Delta T \)

Substitute the values into the formula:
\( q = 225.0\space g \times 4.18\space J/g\cdot^\circ C \times 35.5^\circ C \)

First, multiply \( 225.0 \) and \( 4.18 \):
\( 225.0 \times 4.18 = 940.5 \)

Then, multiply the result by \( 35.5 \):
\( 940.5 \times 35.5 = 940.5 \times 35 + 940.5 \times 0.5 = 32917.5 + 470.25 = 33387.75\space J \approx 3.34\times 10^4\space J \)

Answer:

\( 3.34\times 10^4\space J \) (corresponding to the option \( 3.34 \times 10^4\space J \))