Chapter 3: Q52E (page 577)

A laboratory technician drops a $0.0850 - kg$ sample of unknown solid material, at $100.0 ° C$ , into a calorimeter. The calorimeter can, initially at $19.0 ° C$ , is made of of copper and contains $0.200 kg$ of water .The final temperature of the calorimeter can and contents is $26.1 ° C$ . Compute the specific heat of the sample.

Short Answer

Expert verified

The specific heat of the sample is $1013.3 J / (kg . K)$

Given: Mass of sample is $m_{1} = 0.0850 kg$

Mass of sample is $m_{2} = 0.200 kg$

Mass of sample is $m_{3} = 0.150 kg$

Initial temperature of the calorimeter can (copper and water) is

$T_{i} = T_{2} = T_{3} = 19.0 ° C$

Initial temperature of sample is $T_{1} = 100 ° C$

Final temperature of the system is $T_{f} = 26.1 ° C$

Specific heat of water is $c_{2} = 4.19 \times 10^{3} J / (kg . K)$

Specific heat of copper is $c_{3} = 0.39 \times 10^{3} J / (kg . K)$

Step by step solution

The amount of heat gain by body

The amount of heat gain by body of mass $m$ is given by

$Q = mc ∆ T$

Where , Q is the amount of heat gain by body, is the mass of body, c is the specific heat capacity of body and $∆ T$ is the change in temperature.

The conservation of heat energy

The heat gain by one body is equals to the heat lost by another body.

$Q_{Lost} = Q_{gain}$

Calculation of Specific heat capacity of sample

Heat lost by sample is equals to Heat gain by Calorimeter

$- Q_{sample} = Q_{calorimeter} - Q_{sample} = - Q_{copper} + - Q_{water} - Q_{1} = - Q_{2} + - Q_{3} m_{1} c_{1} ∆ T = - (m_{2} c_{2} + m_{3} + c_{3} ∆ T_{3}) c_{1} = \frac{- (m_{2} c_{2} + m_{3} c_{3}) ∆ T}{m_{1} ∆ T_{1_{}}} c_{1} = \frac{- ((0.150 kg \times 4.19 \times 10^{3} J / (kg . k) - 0.200 kg \times 0.39 \times 10^{3} J / (kg . k)) (26.1 ° C - 19.0 ° C))}{0.0850 (26.1 ° C - 100 ° C)}$

$c_{1} = 1013.3 / (kg . K)$

Thus, the specific heat of the sample is $1013.3 J / (kg . K)$ $1013.3 J / (kg . k)$

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Short Answer

Step by step solution

The amount of heat gain by body

The conservation of heat energy

Calculation of Specific heat capacity of sample

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