Question

2300. Grams of warm water at 30 K is mixed with a solid piece of marble at 80 K. if the temperature settles at 70 K , what must the mass of the piece of marble been? Cp water is 4.18J/gK

Answer 1

The mass of the piece of marble, given that the temperature of the marble water mixture settles at 70 K is 43700 grams

How to calculate the mass of the marble?

Let's begin by calculating the energy release by the marble into the water. Details below:

• Mass of water (M) = 2300 grams
• Initial temperature (T₁) = 30 K
• Final temperature (T₂) = 70 K
• Change in temperature (ΔT) = 70 - 30 = 40 K
• Specific heat capacity of water (C) = 4.18 J/gK
• Energy released by marble (Q) =?

Energy released = - energy absorbed by water

= -(MCΔT)

= -(2300 × 4.18 × 40)

= -384560 J

Now, we shall determine the mass of the marble. Details below:

• Heat released by marble (Q) = -384560 J
• Initial temperature of marble (T₁) = 80 K
• Final temperature of marble (T₂) = 70 K
• Change in temperature (ΔT) = 70 - 80 = -10 K
• Specific heat capacity of marble (C) = 0.880 J/gK
• Mass of marble (M) =?

`M = (Q)/(C\Delta T) \\\\M = (-384560)/(0.880\ *\ -10) \\\\M = 43700\ grams`