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

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:

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

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:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$       ......(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:

$59.1\times 4.186\times (32.0-20.2)=-[24.7\times c_2\times (32.0-96.7)]$

$c_1=1.83J/g^oC$

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