Question

3.4 measuring energy changes: calorimetry
a water sample at 50.0 °c is mixed with 50 g of water at 25.0 °c, and the final temperature becomes 41.67 °c. what was the mass of the warmer sample of water? specific heat capacity = 4.18 j/g°c.
o 100 g
o 50 g
o 10 g
o 5 g

Explanation:

Step1: Set up heat - transfer equation

According to the principle of conservation of energy, the heat lost by the warmer water is equal to the heat gained by the cooler water. The heat - transfer formula is $q = mc\Delta T$. Let the mass of the warmer water be $m_1$. The mass of the cooler water $m_2=50\ g$, the specific heat capacity $c = 4.18\ J/g^{\circ}C$, the initial temperature of the warmer water $T_{1i}=50.0^{\circ}C$, the initial temperature of the cooler water $T_{2i}=25.0^{\circ}C$, and the final temperature $T_f = 41.67^{\circ}C$. The heat lost by the warmer water $q_1=m_1c(T_{1i}-T_f)$ and the heat gained by the cooler water $q_2=m_2c(T_f - T_{2i})$. Since $q_1 = q_2$, we have $m_1c(T_{1i}-T_f)=m_2c(T_f - T_{2i})$. The specific - heat capacity $c$ cancels out on both sides of the equation, and we get $m_1(T_{1i}-T_f)=m_2(T_f - T_{2i})$.

Step2: Substitute the values and solve for $m_1$

Substitute the given values: $T_{1i}=50.0^{\circ}C$, $T_{2i}=25.0^{\circ}C$, $T_f = 41.67^{\circ}C$, and $m_2 = 50\ g$ into the equation $m_1(T_{1i}-T_f)=m_2(T_f - T_{2i})$.
$m_1(50.0 - 41.67)=50(41.67 - 25.0)$
$m_1\times8.33=50\times16.67$
$m_1=\frac{50\times16.67}{8.33}$
$m_1 = 100\ g$

Answer:

A. 100 g