Respuesta :
Answer: The specific heat of metal is 2.34 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 g/mL
Volume of water = 35 mL
Putting values in above equation, we get:
[tex]1g/mL=\frac{\text{Mass of water}}{35mL}\\\\\text{Mass of water}=(1g/mL\times 35mL)=35g[/tex]
When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed 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 metal = 4.82 g
[tex]m_2[/tex] = mass of water = 35 g
[tex]T_{final}[/tex] = final temperature = 34.5°C
[tex]T_1[/tex] = initial temperature of metal = 115°C
[tex]T_2[/tex] = initial temperature of water = 28.7°C
[tex]c_1[/tex] = specific heat of metal = ?
[tex]c_2[/tex] = specific heat of water = 4.186 J/g°C
Putting values in equation 1, we get:
[tex]4.82\times c_1\times (34.5-110)=-[35\times 4.186\times (34.5-28.7)][/tex]
[tex]c_1=2.34J/g^oC[/tex]
Hence, the specific heat of metal is 2.34 J/g°C
