# Question cb561

Feb 27, 2017

${\text{0.75 J g"^(-1)""^@"C}}^{- 1}$

#### Explanation:

The idea here is that the heat given off by the metal as it cooled from ${84}^{\circ} \text{C}$ to ${36.5}^{\circ} \text{C}$ is equal to the heat absorbed by the water as it warmed from ${29}^{\circ} \text{C}$ to ${36.5}^{\circ} \text{C}$.

color(blue)(ul(color(black)(-q_"metal" = q_"water")))" " " "color(red)("(*)")

The minus sign is used here because, by definition, heat given off carries a negative sign.

So, you should know that water has a specific heat of about

${c}_{\text{water"= "4.18 j g"^(-1)""^@"C}}^{- 1}$

You can use the information provided by the problem to figure out how much heat was absorbed by the water.

Your tool of choice here will be the equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{q = m \cdot c \cdot \Delta T}}}$

Here

• $q$ is the heat lost or gained by the substance
• $m$ is the mass of the sample
• $c$ is the specific heat of the substance
• $\Delta T$ is the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample

In your case, you know that the change in temperature for the water is equal to

$\Delta T = {36.5}^{\circ} \text{C" - 29^@"C" = 7.5^@"C}$

which means that you have

q_"water" = 41 color(red)(cancel(color(black)("g"))) * "4.18 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 7.5color(red)(cancel(color(black)(""^@"C")))

${q}_{\text{water" = "1285.35 J}}$

Now, according to equation $\textcolor{red}{\text{(*)}}$, this is exactly how much heat was given off by the metal. You can thus say that

$- {q}_{\text{metal" = "1285.35 J" implies q_"metal" = - "1285.35 J}}$

Use the same equation to find the specific heat of the metal, ${c}_{\text{metal}}$. Rearrange the equation to solve for ${c}_{\text{metal}}$

c_"metal" = (q_"metal")/(m_"metal" * DeltaT_"metal")

In this case, you have

$\Delta {T}_{\text{metal" = 36.5^@"C" - 84^@"C" = -46.5^@"C}}$

This means that the metal has a specific heat equal to

c_"metal" = (color(red)(cancel(color(black)(-)))"1285.35 J")/(37color(red)(cancel(color(black)("g"))) * (color(red)(cancel(color(black)(-)))46.5^@"C")) = color(darkgreen)(ul(color(black)("0.75 J g"^(-1)""^@"C"^(-1))))#

The answer is rounded to two sig figs.