A24.7 g sample of beryllium at 96.7°c is placed into 59.1 ml of water at 20.2°c in an insulated container. the temperature of the water at thermal equilibrium is 32.0°c. what is the specific heat of beryllium? assume a density of 1.00 g/ml for water

The specific heat of beryllium is 1.83 J/g°C

Explanation:

To calculate the mass of water, we use the equation:

Density of water = 1.00 g/mL

Volume of water = 59.1 mL

Putting values in above equation, we get:

When berrylium is dipped in water, the amount of heat absorbed by metal will be equal to the amount of heat released by water.

The equation used to calculate heat released or absorbed follows:

      ......(1)

where,

q = heat absorbed or released

= mass of water = 24.7 g

= mass of beryllium= 59.1 g

= final temperature = 32.0°C

= initial temperature of water = 20.2°C

= initial temperature of  beryllium =96.7°C

= specific heat of water= 4.186 J/g°C

= specific heat of beryllium = ?

Putting values in equation 1, we get:

Hence, the specific heat of beryllium is 1.83 J/g°C

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explanation:

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