Question

You have 80 g of 60oC water and 80 g or 10oC water. What is the final temperature when the two are mixed? Explain.

Answer 1

Answer:

35 C

Explanation:

At thermal equilibrium, the heat given by the hotter mass of water must be equal to the heat absorbed by the colder water (because energy cannot be destroyed nor created). In formula:

$Q_h = -Q_c\\m_h C_h \Delta T_h = -m_c C_c \Delta T_c$

where:

$m_h = 80 g$ is the mass of the hot water

Ch is the specific heat of the hot water

$\Delta T_h$ is the variation in temperature of the hot water

$m_c = 80 g$ is the mass of the cold water

Cc is the specific heat of the cold water

$\Delta T_c$ is the variation in temperature of the c water

We notice that:

$m_h = m_c$ since the two masses are equal, and

$C_h = C_c$ because the substance is the same (water)

So, the above equation just simplifies as

$\Delta T_h =- \Delta T_c$

which can be rewritten as:

$T_f - 60 C = -(T_f - 10 C)$

where T_f is the temperature of both masses of water at equilibrium. By solving the equation, we find

$2 T_f = 70 C\\T_f = 35 C$

Answer 2

The answer is 35 degrees Celsius