   Chapter 11, Problem 59AP

Chapter
Section
Textbook Problem

A student measures the following data in a calorimetry experiment designed to determine the specific heat of aluminum: Initial temperature of water and calorimeter: 70.0°C Mass of water: 0.400 kg Mass of calorimeter: 0.040 kg Specific heat of calorimeter: 0.63 kJ/kg · °C Initial temperature of aluminum: 27.0°C Mass of aluminum: 0.200 kg Final temperature of mixture: 66.3°C Use these data to determine the specific heat of aluminum. Explain whether your result is within 15% of the value listed in Table 11.1.

To determine
The specific heat of aluminum.

Explanation

Given Info: Initial temperature of water and calorimeter 70.0°C , Mass of water is 0.400kg , Mass of calorimeter is 0.040 kg, specific heat of calorimeter 0.63kJ/kg°C , Initial temperature of aluminum 27.0°C , mass of aluminum 0.200 kg, and final temperature of mixture is 66.3°C .

Let us consider, Aluminum-Water – Calorimeter systems is a closed system. So, heat gained by Aluminum is equal to the heat lost by the water and calorimeter.

Formula to calculate the volume of the Sun is,

mAcA(TfTi,A)=mwcw(TfTi,w)mccc(TfTi,c)

• mA is mass of aluminum,
• cA is the specific heat of aluminum,
• Ti,A initial temperature of aluminum,
• mw is mass of water,
• cw is the specific heat of water,
• Ti,w is the initial temperature of water,
• mc is the mass of calorimeter,
• cc is the specific heat of calorimeter,
• Ti,c is the initial temperature of calorimeter
• Tf is the final temperature of the mixture.

Rearrange the above equation in terms of cA .

cA=mwcw(TfTi,w)mccc(TfTi,c)mA(TfTi,A)

Substitute 70.0°C for Ti,w , 4186J/kg°C for cw , 70.0°C for Ti,c , 0.400 kg for mw ,

0.040 kg for mc , 0.63kJ/kg°C for cc , 27

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