Answer:
The correct answer is option B.
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]900 J/kg^oC[/tex]
[tex]C_2[/tex] = specific heat of coffee= [tex]4000 J/kg^oC[/tex]
[tex]m_1[/tex] = mass of metal = x
[tex]m_2[/tex] = mass of coffee = 0.3 kg
[tex]T_f[/tex] = final temperature of aluminum metal= [tex]110^oC[/tex]
[tex]T_1[/tex] = initial temperature of aluminum metal = [tex]-10^oC[/tex]
[tex]T_2[/tex] = initial temperature of coffee= [tex]140^oC[/tex]
Now put all the given values in the above formula, we get
[tex]x\times 900 J/kg^oC\times (110-(-10))^oC=-(0.3 kg\times 4000 J/kg^oC\times (110-140)^oC[/tex]
[tex]x=0.333 kg[/tex]
Mass of aluminum cubes = 0.3333 kg = 333.3 g
If mass of 1 cube is 1 gram, then numbers of cubes in 333.3 grams will be:
[tex]=\frac{333.3 g}{1 g}=333.3\approx 330[/tex]
330 cubes of aluminum cubes will be required.
