Answer : The mass of the water in two significant figures is, [tex]3.0\times 10^1g[/tex]
Explanation :
In this case the 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 iron metal = [tex]0.45J/g^oC[/tex]
[tex]c_2[/tex] = specific heat of water = [tex]4.18J/g^oC[/tex]
[tex]m_1[/tex] = mass of iron metal = 32.3 g
[tex]m_2[/tex] = mass of water = ?
[tex]T_f[/tex] = final temperature of mixture = [tex]59.2^oC[/tex]
[tex]T_1[/tex] = initial temperature of iron metal = [tex]21.9^oC[/tex]
[tex]T_2[/tex] = initial temperature of water = [tex]63.5^oC[/tex]
Now put all the given values in the above formula, we get
[tex]32.3g\times 0.45J/g^oC\times (59.2-21.9)^oC=-m_2\times 4.18J/g^oC\times (59.2-63.5)^oC[/tex]
[tex]m_2=30.16g\approx 3.0\times 10^1g[/tex]
Therefore, the mass of the water in two significant figures is, [tex]3.0\times 10^1g[/tex]
The mass of the water in the insulated container is 30.04 g
Heat loss = Heat gain
MᵥᵥC(Tᵥᵥ – Tₑ) = MᵢC(Tₑ – Mᵢ)
Mᵥᵥ × 4.184 (63.5 – 59.2) = 32.2 × 0.45(59.2 – 21.9)
Mᵥᵥ × 4.184(4.3) = 14.49(37.3)
Clear bracket
Mᵥᵥ × 17.9912 = 540.477
Divide both side by 17.9912
Mᵥᵥ = 540.477 / 17.9912
Mᵥᵥ = 30.04 g
Learn more about heat transfer:
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