Question

A 24.7 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

Answer 1

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

Step-by-step explanation:

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

`\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}`

Density of water = 1.00 g/mL

Volume of water = 59.1 mL

Putting values in above equation, we get:

`1.00g/mL=\frac{\text{Mass of water}}{100.0mL}\\\\\text{Mass of water}=(1.00g/mL* 59.1mL)=59.1g`

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

`Heat_{\text{absorbed}}=Heat_{\text{released}}`

The equation used to calculate heat released or absorbed follows:

`Q=m* c* \Delta T=m* c* (T_(final)-T_(initial))`

`m_1* c_1* (T_(final)-T_1)=-[m_2* c_2* (T_(final)-T_2)]` ......(1)

where,

q = heat absorbed or released

`m_1` = mass of water = 24.7 g

`m_2` = mass of beryllium= 59.1 g

`T_(final)` = final temperature = 32.0°C

`T_1` = initial temperature of water = 20.2°C

`T_2` = initial temperature of beryllium =96.7°C

`c_1` = specific heat of water= 4.186 J/g°C

`c_2` = specific heat of beryllium = ?

Putting values in equation 1, we get:

`59.1* 4.186* (32.0-20.2)=-[24.7* c_2* (32.0-96.7)]`

`c_1=1.83J/g^oC`

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