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3 years ago

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A laboratory technician drops a 72.0 g sample of unknown solid material, at a temperature of 80.0°C, into a calorimeter. The cal

orimeter can, initially at 11.0°C, is made of 187 g of copper and contains 203 g of water. The final temperature of the calorimeter can and contents is 39.4°C. What is the specific heat of the unknown sample? Please give your answer in units of J/kg.°C.

1 answer:

Natalija [7]3 years ago

7 0

Answer : The specific heat of unknown sample is, $8748.78J/kg^oC$

Explanation :

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

$q_1=-[q_2+q_3]$

$m_1\times c_1\times (T_f-T_1)=-[m_2\times c_2\times (T_f-T_2)+m_3\times c_3\times (T_f-T_2)]$

where,

$c_1$ = specific heat of unknown sample = ?

$c_2$ = specific heat of water = $4186J/kg^oC$

$c_3$ = specific heat of copper = $390J/kg^oC$

$m_1$ = mass of unknown sample = 72.0 g = 0.072 kg

$m_2$ = mass of water = 203 g = 0.203 kg

$m_2$ = mass of copper = 187 g = 0.187 kg

$T_f$ = final temperature of calorimeter = $39.4^oC$

$T_1$ = initial temperature of unknown sample = $80.0^oC$

$T_2$ = initial temperature of water and copper = $11.0^oC$

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

$0.072kg\times c_1\times (39.4-80.0)^oC=-[(0.203kg\times 4186J/kg^oC\times (39.4-11.0)^oC)+(0.187kg\times 390J/kg^oC\times (39.4-11.0)^oC)]$

$c_1=8748.78J/kg^oC$

Therefore, the specific heat of unknown sample is, $8748.78J/kg^oC$

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