A 30-0 g sample of water at 280 K is mixed with 50.0 g of water at 330 K. How would you calculate the final temperature of the mixture assuming no heat is lost to the surroundings?

Question

Cameron Andrews · Accepted Answer

#"310 K"#

It is possible to say that the heat lost by the hotter sample will equal the heat absorbed by the cooler sample if you assume that no heat is lost to the environment.

This is the equation that now shows a connection between heat gained or lost and temperature change.

#color(blue)(q = m * c * DeltaT)" "#, where

#q# - heat lost or gained #m# - the mass of the sample #c# - the specific heat of the substance #DeltaT# - the change in temperature, defined as final temperature minus initial temperature

Now, try to anticipate what will happen when the two samples are combined before performing any calculations.

Notice that you have more hot water than cold water, which means that you can expect the final temperature of the mixture to be closer to #"330 K"# than to #"280 K"#.

Thus, one could say that

#-q_"lost" = q_"gained"#

Since heat loss is negative, the minus sign is used in this instance.

This is the same as

#m_"hot" * color(red)(cancel(color(black)(c))) * DeltaT_"hot" = m_"cold" * color(red)(cancel(color(black)(c))) * DeltaT_"cold"#

Let's say that the final temperature is #color(blue)(T_"f")#. This means that you have

#-50.0color(red)(cancel(color(black)("g"))) * (color(blue)(T_"f") - 330)color(red)(cancel(color(black)("K"))) = 30.0color(red)(cancel(color(black)("g"))) * (color(blue)(T_"f") - 280)color(red)(cancel(color(black)("K")))#

#-50.0 * color(blue)(T_"f") + 16500 = 30.0 * color(blue)(T_"f") - 8400#

#80.0 * color(blue)(T_"f") = 24900#

#color(blue)(T_"f") = 24900/80.0 = "311.25 K"#

The number of sig figs you have for the two temperatures, rounded to two sig figs, is what the answer will be.

#T_"f" = color(green)("310 K")#

Serenity Allen · Answer

undefined Use the principle of conservation of energy.

$$ m_1c_1(T_f - T_1) + m_2c_2(T_f - T_2) = 0 $$

Solve for $ T_f $:

$$ T_f = \frac{m_1c_1T_1 + m_2c_2T_2}{m_1c_1 + m_2c_2} $$

where:
$ m_1 = 30.0 \, \text{g} $ (mass of first water sample)
$ c_1 = 4.18 \, \text{J/g} \cdot \text{K} $ (specific heat of water)
$ T_1 = 280 \, \text{K} $ (initial temperature of first water sample)
$ m_2 = 50.0 \, \text{g} $ (mass of second water sample)
$ c_2 = 4.18 \, \text{J/g} \cdot \text{K} $ (specific heat of water)
$ T_2 = 330 \, \text{K} $ (initial temperature of second water sample)