Question

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?

Answer 1

T₂ = 16.83°C

Step-by-step explanation:

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

`Energy\ Lost\ by\ Metal = Energy\ Gaine\ by\ Water\\m_(metal)C_(metal)(T_2-T_(1metal)) = m_(w)C_(w)(T_2-T_(1w))`

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
`(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\\` 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,

`(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\\`

T₂ = 16.83°C