Question

zoe has 25 grams of water (c = 4.186 \\(\frac{j}{g ^\circ c}\\)) at 10\\(^\circ\\)c, which she mixes with 12 grams of water at 30\\(^\circ\\)c. assume that no heat is lost to the surroundings. what is the final temperature of the two liquids? the answer is expressed to the nearest whole number. \\(\circ\\) -47\\(^\circ\\)c \\(\circ\\) -8\\(^\circ\\)c \\(\circ\\) 3\\(^\circ\\)c \\(\circ\\) 16\\(^\circ\\)c

Explanation:

Step1: Define Heat Transfer

When mixing two samples of water (same substance, so same specific heat \( c \)), the heat gained by the colder water equals the heat lost by the warmer water. Let \( T_f \) be the final temperature. The formula for heat is \( q = mc\Delta T \), where \( m \) is mass, \( c \) is specific heat, and \( \Delta T \) is temperature change.

For the colder water (25 g, \( 10^\circ\text{C} \)): \( q_{\text{gained}} = m_1c(T_f - T_1) \)
For the warmer water (12 g, \( 30^\circ\text{C} \)): \( q_{\text{lost}} = m_2c(T_2 - T_f) \)

Since \( q_{\text{gained}} = q_{\text{lost}} \) (no heat loss), \( m_1c(T_f - T_1) = m_2c(T_2 - T_f) \). The \( c \) cancels out.

Step2: Substitute Values

\( m_1 = 25 \, \text{g} \), \( T_1 = 10^\circ\text{C} \); \( m_2 = 12 \, \text{g} \), \( T_2 = 30^\circ\text{C} \).

\( 25(T_f - 10) = 12(30 - T_f) \)

Step3: Solve for \( T_f \)

Expand both sides:
\( 25T_f - 250 = 360 - 12T_f \)

Add \( 12T_f \) to both sides:
\( 37T_f - 250 = 360 \)

Add 250 to both sides:
\( 37T_f = 610 \)

Divide by 37:
\( T_f = \frac{610}{37} \approx 16.49^\circ\text{C} \), which rounds to \( 16^\circ\text{C} \).

Answer:

D. \( 16^\circ\text{C} \) (assuming the options are labeled with D for the last option; if options are A-D as: A. \(-47^\circ\text{C}\), B. \(-8^\circ\text{C}\), C. \(3^\circ\text{C}\), D. \(16^\circ\text{C}\), then the answer is D. \( 16^\circ\text{C} \))