Question

Fifty-four grams of a certain metal at 98°C was placed in 80 mL of water qt 297 K. Assuming no heat is lost to the surroundings, what is the temperature of water and the metal? (Specific heat of the metal = 0.085 cal/g*C°)

Answer 1

Answer:

297K. Assuming no heat is lost to the surrounding, what is the temperature of the water and the metal? (Specific heat of metal = 0.085cal/g degree Celsius) ...

Explanation:

Answer 2

The temperature of water and the metal is 27.3°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 g/mL

Volume of water = 80 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{80mL}\\\\\text{Mass of water}=(1g/mL\times 80mL)=80g$

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed 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 metal = 44 g

$m_2$ = mass of water = 80 g

$T_{final}$ = final temperature = ?°C

$T_1$ = initial temperature of metal = 98°C

= initial temperature of water = 297 K = [297 - 273] = 24

°C

$c_1$ = specific heat of metal = 0.085 Cal/g°C

$c_2$ = specific heat of water= 1 Cal/g°C

Putting values in equation 1, we get:

$44\times 0.085\times (T_{final}-98)=-[80\times 1\times (T_{final}-24)]$

$T_{final}=27.3^oC$

Learn more about thermal equilibrium:

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