A student wishes to determine the heat capacity of a coffee-cup calorimeter. After she mixes 108.7 g of water at 60.2°C with 108.7 g of water, already in the calorimeter, at 19.3°C, the final temperature of the water is 35.0°C. Calculate the heat capacity of the calorimeter in J/K. Use 4.184 J/g°C as the specific heat of water. Enter to 1 decimal place.

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Alleei Alleei · Answer 1 · 2020-02-20T11:31:44+01:00

Answer : The heat capacity of the calorimeter is, [tex]6.72J/g^oC[/tex]

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

[tex]q_1=-q_2[/tex]

[tex]m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)[/tex]

where,

[tex]c_1[/tex] = specific heat of water = [tex]4.184J/g^oC[/tex]

[tex]c_2[/tex] = specific heat of calorimeter = ?

[tex]m_1[/tex] = mass of water = 108.7 g

[tex]m_2[/tex] = mass of calorimeter = 108.7 g

[tex]T_f[/tex] = final temperature of mixture = [tex]35.0^oC[/tex]

[tex]T_1[/tex] = initial temperature of water = [tex]60.2^oC[/tex]

[tex]T_2[/tex] = initial temperature of calorimeter = [tex]19.3^oC[/tex]

Now put all the given values in the above formula, we get

[tex](108.7g)\times (4.184J/g^oC)\times (35.0-60.2)^oC=-(108.7g)\times c_2\times (35.0-19.3)^oC[/tex]

[tex]c_2=6.72J/g^oC[/tex]

Therefore, the heat capacity of the calorimeter is, [tex]6.72J/g^oC[/tex]