Question

A 333.7g sample is heated to 98.7°C and placed into a calorimeter containing 72.4 g of water at 23.60°C. The final temperature in the calorimeter was 32.40°C. What is the specific heat of the mineral?

Answer 1

To calculate the specific heat of the mineral, we can use the equation:

Q = m × c × ΔT

Where: Q is the heat absorbed or released m is the mass c is the specific heat ΔT is the change in temperature

First, let's calculate the heat absorbed by the water:

Q_water = m_water × c_water × ΔT_water

Given: m_water = 72.4 g c_water (specific heat of water) = 4.18 J/g°C ΔT_water = (final temperature - initial temperature) = (32.40°C - 23.60°C) = 8.80°C

Q_water = 72.4 g × 4.18 J/g°C × 8.80°C

Now, let's calculate the heat released by the mineral:

Q_mineral = m_mineral × c_mineral × ΔT_mineral

Given: m_mineral = 333.7 g c_mineral (specific heat of the mineral) = ? ΔT_mineral = (final temperature - initial temperature) = (32.40°C - 98.7°C) = -66.30°C (negative because heat is released)

Now, we can equate the two equations and solve for c_mineral:

Q_water = -Q_mineral

72.4 g × 4.18 J/g°C × 8.80°C = -(333.7 g × c_mineral × -66.30°C)

Simplifying:

72.4 g × 4.18 J/g°C × 8.80°C = 333.7 g × c_mineral × 66.30°C

Now, solve for c_mineral:

c_mineral = (72.4 g × 4.18 J/g°C × 8.80°C) / (333.7 g × 66.30°C)

c_mineral = 0.712 J/g°C

Therefore, the specific heat of the mineral is approximately 0.712 J/g°C.