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HELP!!!!

A student did an experiment to determine the
specific heat capacity of an unknown metal.
She heated 1.00 x 10- kg of the metal to 225°C
and quickly placed it in an insulated container
(negligible specific heat capacity) that contained
0.0900 kg of water at a temperature of 18.0°C.
What is the final temperature of the water if the
specific heat capacity of the metal is
2.11 x 102 J/kg.°C?

Respuesta :

hamzaahmeds

Answer:

T₂ = 16.83°C

Explanation:

Applying the law of conservation of energy principle here in this situation we get the following equation:

[tex]Energy\ Lost\ by\ Metal = Energy\ Gaine\ by\ Water\\m_{metal}C_{metal}(T_2-T_{1metal}) = m_{w}C_{w}(T_2-T_{1w})[/tex]

where,

T₂ = Final Temperature of Water = Final Temperature of Metal = ?

C_metal = Specififc Heat Capacity of the metal = 2.11 x 10² J/lg.°C

T_1metal = Initial Temperature of Metal = 225°C

m_metal = mass of metal = 1 x 10⁻²[tex](0.01\ kg)(211\ J/kg.^oC)(T_2-225^oC) = (0.09\ kg)(4184\ J/kg.^oC)(T_2-18^oC)\\2.11 T_2 - 474.75 = 376.56T_2 - 6778.08\\374.45T_2 = 6303.33\\[/tex] kg (exponent assumed due to missing info in question)

C_w = Specififc Heat Capacity of the water = 4184 J/lg.°C

T_1w = Initial Temperature of water = 18°C

m_w = mass of water = 0.09 kg

Therefore,

[tex](0.01\ kg)(211\ J/kg.^oC)(T_2-225^oC)=(0.09\ kg)(4184\ J/kg.^oC)(T_2-18^oC)\\\\2.11 - 474.75T_2 = 376.56 - 6778.08T_2\\[/tex]

T₂ = 16.83°C

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