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A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimeter containing 75 g of water at 20°C.
The calorimeter is constructed of a material that has a specific heat of 0.10 cal/ g⋅°C.

When equilibrium is reached, what will be the final temperature? cwater = 1.00 cal/g⋅°C.

a. 114°Cb. 72°Cc. 64°Cd. 37°C

Respuesta :

Answer:

d. 37 °C

Explanation:

[tex]m_{m}[/tex] = mass of lump of metal = 250 g

[tex]c_{m}[/tex] = specific heat of lump of metal  = 0.25 cal/g°C

[tex]T_{mi}[/tex] = Initial temperature of lump of metal = 70 °C

[tex]m_{w}[/tex] = mass of water = 75 g

[tex]c_{w}[/tex] = specific heat of water = 1 cal/g°C

[tex]T_{wi}[/tex] = Initial temperature of water = 20 °C

[tex]m_{c}[/tex] = mass of calorimeter  = 500 g

[tex]c_{c}[/tex] = specific heat of calorimeter = 0.10 cal/g°C

[tex]T_{ci}[/tex] = Initial temperature of calorimeter = 20 °C

[tex]T_{f}[/tex] = Final equilibrium temperature

Using conservation of heat

Heat lost by lump of metal = heat gained by water + heat gained by calorimeter

[tex]m_{m} c_{m} (T_{mi} - T_{f}) = m_{w} c_{w} (T_{f} - T_{wi}) +  m_{c} c_{c} (T_{f} - T_{ci}) \\(250) (0.25) (70 - T_{f} ) = (75) (1) (T_{f} - 20) + (500) (0.10) (T_{f} - 20)\\T_{f} = 37 C[/tex]

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