Question

zoe has 25 grams of water (c = 4.186 j/(g°c)) at 10°c, which she mixes with 12 grams of water at 30°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. - 47°c - 8°c - 3°c - 16°c

Explanation:

Step1: Set up heat - transfer equation

Since no heat is lost to the surroundings, $Q_{lost}=Q_{gained}$. The heat - transfer formula is $Q = mc\Delta T$. Let the final temperature be $T$. For the cooler water ($m_1 = 25\ g$, $T_1=10^{\circ}C$) and the warmer water ($m_2 = 12\ g$, $T_2 = 30^{\circ}C$), and since $c$ is the same for water in both cases ($c = 4.186\ J/(g^{\circ}C)$), we have $m_1c(T - T_1)=m_2c(T_2 - T)$. The $c$ cancels out, and we get $m_1(T - T_1)=m_2(T_2 - T)$.

Step2: Expand the equation

$25(T - 10)=12(30 - T)$. Expand to get $25T-250 = 360-12T$.

Step3: Solve for $T$

Add $12T$ to both sides: $25T + 12T-250=360$, which simplifies to $37T-250 = 360$. Then add 250 to both sides: $37T=360 + 250=610$. So, $T=\frac{610}{37}\approx16.49^{\circ}C$. Rounding to the nearest whole number gives $T = 16^{\circ}C$.

Answer:

D. 16°C