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a 0.350 kg block at -27.5 °C is added to 0.217 kg of water at 25.0 °C. they come to equilibrium at 16.4 °C. what is the specific heat of the block?

a 0350 kg block at 275 C is added to 0217 kg of water at 250 C they come to equilibrium at 164 C what is the specific heat of the block class=

Respuesta :

Answer:

[tex]C_{pb}=0.501\ kJ/kg.K[/tex]

Explanation:

Given that

[tex]m_1=0.35 kg[/tex]

[tex]T_1=-27.5^oC[/tex]

[tex]m_2=0.214 kg[/tex]

[tex]T_2=25^oC[/tex]

[tex]T=16.4^oC[/tex]

We know that

[tex]C_{pw}=4.187 kJ/kg.K[/tex]

By using energy conservation

Heat lost by water = Heat gain by block

[tex]m_2\times C_{pw}\times (T_2-T)=m_1\times C_{pb}\times (T-T_1)[/tex]

[tex]0.214\times 4.187\times (25-16.4)=0.35\times C_{pb}\times (16.4+27.5)[/tex]

[tex]C_{pb}=0.501\ kJ/kg.K[/tex]

Therefore the specific heat of the block will be 0.501 kJ/kg.K

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