Answer :
Answer: The specific heat of beryllium is 1.83 J/g°C
Explanation:
To calculate the mass of water, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
Density of water = 1.00 g/mL
Volume of water = 59.1 mL
Putting values in above equation, we get:
[tex]1.00g/mL=\frac{\text{Mass of water}}{100.0mL}\\\\\text{Mass of water}=(1.00g/mL\times 59.1mL)=59.1g[/tex]
When berrylium is dipped in water, the amount of heat absorbed by metal will be equal to the amount of heat released by water.
[tex]Heat_{\text{absorbed}}=Heat_{\text{released}}[/tex]
The equation used to calculate heat released or absorbed follows:
[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]
[tex]m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)][/tex] ......(1)
where,
q = heat absorbed or released
[tex]m_1[/tex] = mass of water = 24.7 g
[tex]m_2[/tex] = mass of beryllium= 59.1 g
[tex]T_{final}[/tex] = final temperature = 32.0°C
[tex]T_1[/tex] = initial temperature of water = 20.2°C
[tex]T_2[/tex] = initial temperature of beryllium =96.7°C
[tex]c_1[/tex] = specific heat of water= 4.186 J/g°C
[tex]c_2[/tex] = specific heat of beryllium = ?
Putting values in equation 1, we get:
[tex]59.1\times 4.186\times (32.0-20.2)=-[24.7\times c_2\times (32.0-96.7)][/tex]
[tex]c_1=1.83J/g^oC[/tex]
Hence, the specific heat of beryllium is 1.83 J/g°C
Final answer:
The estimated specific heat of beryllium based on this experiment is 1.83 J/g°C.
Explanation:
To determine the specific heat of beryllium, we apply the concept that the heat lost by the beryllium equals the heat gained by the water (assuming a perfectly insulated system and no heat loss to the surroundings). The equation for heat transfer is ΔQ = mcΔT, where ΔQ is the heat transferred, m is the mass, c is the specific heat, and ΔT is the temperature change.
For water, ΔQwater = mwatercwaterΔTwater, with mwater = 59.1 g (given that 1.00 mL = 1.00 g), cwater = 4.186 J/g°C, and ΔTwater = 32.0 °C - 20.2 °C = 11.8 °C.
For beryllium, ΔQberyllium = -ΔQwater because the heat lost by beryllium is gained by water. Thus, mberylliumcberylliumΔTberyllium = -mwatercwaterΔTwater, with mberyllium = 24.7 g, and ΔTberyllium = 32.0 °C - 96.7 °C = -64.7 °C.
Solving for cberyllium and substituting the known values, we find that the specific heat of beryllium can be calculated as cberyllium = -(mwatercwaterΔTwater) / (mberylliumΔTberyllium) = 59.1 g * 4.186 J/g°C * 11.8 °C) / (24.7 g * -64.7 °C.
cberyllium ≈ 1.83 J/g°C.
