Question

if you combine 420.0 ml of water at 25.00°c and 100.0 ml of water at 95.00°c, what is the final temperature of the mixture? use 1.00 g/ml as the density of water.

t_final = °c

Explanation:

Step1: Calculate the mass of each water sample

The density of water is \(1.00\ g/mL\).
For the first sample: \(m_1 =
ho V_1=1.00\ g/mL\times420.0\ mL = 420.0\ g\)
For the second sample: \(m_2=
ho V_2 = 1.00\ g/mL\times100.0\ mL=100.0\ g\)

Step2: Use the heat - transfer formula \(q = mc\Delta T\) (where \(c\) is the specific heat capacity of water, \(c = 4.184\ J/(g\cdot^{\circ}C)\))

Since \(q_{lost}=q_{gained}\) (assuming no heat loss to the surroundings), \(m_1c(T_{final}-T_{1})=m_2c(T_{2}-T_{final})\)
Cancel out \(c\) (because \(c\) is the same for water)
\(m_1(T_{final}-T_{1})=m_2(T_{2}-T_{final})\)

Step3: Substitute the values of \(m_1\), \(m_2\), \(T_1\) and \(T_2\)

\(420.0(T_{final}- 25.00)=100.0(95.00 - T_{final})\)
Expand both sides:
\(420.0T_{final}-420.0\times25.00=100.0\times95.00 - 100.0T_{final}\)
\(420.0T_{final}-10500 = 9500-100.0T_{final}\)

Step4: Rearrange the equation to solve for \(T_{final}\)

\(420.0T_{final}+ 100.0T_{final}=9500 + 10500\)
\(520.0T_{final}=20000\)
\(T_{final}=\frac{20000}{520.0}\approx38.5^{\circ}C\)

Answer:

\(38.5^{\circ}C\)