RissaB03
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. 125g of water has an initial temperature of 25.6°C, and is heated by 50.0g of a metal
which has been heated to 115.0°C. The metal heats the water so that both the metal
and the water reach a final temperature of 29.3°C. Calculate the specific heat of the
metal.​

125g of water has an initial temperature of 256C and is heated by 500g of a metalwhich has been heated to 1150C The metal heats the water so that both the meta class=

Respuesta :

Answer : The specific heat of the metal is, [tex]0.451J/g^oC[/tex]

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

[tex]q_1=-q_2[/tex]

[tex]m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)[/tex]

where,

[tex]c_1[/tex] = specific heat of metal = ?

[tex]c_2[/tex] = specific heat of water = [tex]4.18J/g^oC[/tex]

[tex]m_1[/tex] = mass of metal = 50.0 g

[tex]m_2[/tex] = mass of water  = 125 g

[tex]T_f[/tex] = final temperature of mixture = [tex]29.3^oC[/tex]

[tex]T_1[/tex] = initial temperature of metal = [tex]115.0^oC[/tex]

[tex]T_2[/tex] = initial temperature of water = [tex]25.6^oC[/tex]

Now put all the given values in the above formula, we get

[tex](50.0g)\times c_1\times (29.3-115.0)^oC=-[(125g)\times 4.18J/g^oC\times (29.3-25.6)^oC][/tex]

[tex]c_1=0.451J/g^oC[/tex]

Therefore, the specific heat of the metal is, [tex]0.451J/g^oC[/tex]