Help me please answer this with solution....A piece of granite weighing 250g is heated in a boiling water to 100°C. When a granite is place in a calorimeter containing 400g water, the temperature of the water increases from 20°C to 28.5°C. What is the specific heat of the granite, assuming all the heat is transferred to the water?

heat lost by granite + heat gained by H2O = 0

[mass granite x specific heat granit x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute and solve for specific heat granite, the only unknown.

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Eduard22sly Eduard22sly · Answer 2 · 2021-11-30T15:04:55+01:00

The specific heat capacity of the granite is 0.796 J/gºC

We'll begin by calculating the heat absorbed by the water. This can be obtained as follow:

Mass of water (M) = 400 g

Initial temperature of water (T₁) = 20 °C

Final temperature (T₂) = 28.5 °C

Change in temperature (ΔT) = T₂ – T₁ = 28.5 – 20 = 8.5 °C

Specific heat capacity of water (C) = 4.184 J/gºC

Heat absorbed (Q) =?

Q = MCΔT

Q = 400 × 4.184 × 8.5

Q = 14225.6 J

Thus, the heat absorbed by the water is 14225.6 J

Finally, we shall determine the specific heat capacity of the granite

Heat absorbed = Heat released

Heat absorbed = 14225.6 J

Heat released = –14225.6 J

Mass of granite (M) = 250 g

Initial temperature of granite (T₁) = 100 °C

Final temperature (T₂) = 28.5 °C

Change in temperature (ΔT) = T₂ – T₁ = 28.5 – 100 = –71.5 °C

Specific heat capacity of granite (C) =?

Q = MCΔT

–14225.6 = 250 × C × –71.5

–14225.6 = –17875 × C

Divide both side by –17875

C = –14225.6 / –17875

C = 0.796 J/gºC

Therefore, the specific heat capacity of the granite is 0.796 J/gºC

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