Question

a 22.0 g sample of brass, which has a specific heat capacity of 0.375 j·g⁻¹·°c⁻¹, is dropped into an insulated container containing 200.0 g of water at 55.8 °c and a constant pressure of 1 atm. the initial temperature of the brass is 3.8 °c.
assuming no heat is absorbed from or by the container, or the surroundings, calculate the equilibrium temperature of the water. be sure your answer has the correct number of significant digits.

Explanation:

Step1: Recall heat transfer formula

The heat gained by the brass (\(q_{brass}\)) is equal to the heat lost by the water (\(q_{water}\)) since the container is insulated. The formula for heat is \(q = mc\Delta T\), where \(m\) is mass, \(c\) is specific heat capacity, and \(\Delta T\) is the change in temperature. Let the equilibrium temperature be \(T\). For brass: \(m_{brass}=22.0\ g\), \(c_{brass}=0.375\ J\ g^{-1}\ ^\circ C^{-1}\), \(\Delta T_{brass}=T - 3.8^\circ C\). For water: \(m_{water}=200.0\ g\), \(c_{water}=4.184\ J\ g^{-1}\ ^\circ C^{-1}\) (specific heat of water), \(\Delta T_{water}=85.0^\circ C - T\). So, \(q_{brass}=m_{brass}c_{brass}(T - 3.8)\) and \(q_{water}=m_{water}c_{water}(85.0 - T)\). Since \(q_{brass}=-q_{water}\) (heat gained by brass is heat lost by water), we have:

$$22.0\times0.375\times(T - 3.8)=-200.0\times4.184\times(85.0 - T)$$

Step2: Simplify the equation

First, calculate the left - hand side coefficient: \(22.0\times0.375 = 8.25\). The right - hand side coefficient: \(200.0\times4.184 = 836.8\). So the equation becomes:

$$8.25(T - 3.8)=-836.8(85.0 - T)$$

Expand both sides:

$$8.25T-8.25\times3.8=-836.8\times85.0 + 836.8T$$

Calculate \(8.25\times3.8 = 31.35\) and \(836.8\times85.0=71128\). So:

$$8.25T - 31.35=-71128+836.8T$$

Step3: Rearrange terms to solve for T

Move all terms with \(T\) to one side and constants to the other side:

$$8.25T-836.8T=-71128 + 31.35$$

$$ - 828.55T=-71096.65$$

Then, solve for \(T\):

$$T=\frac{71096.65}{828.55}\approx85.8^\circ C$$

Wait, no, we made a sign error in the heat transfer. The correct relationship is \(q_{brass}=-q_{water}\), which means \(m_{brass}c_{brass}(T - T_{brass})=m_{water}c_{water}(T_{water}-T)\) (because brass is at lower temperature, it gains heat, water is at higher temperature, it loses heat). So the correct equation is:

$$22.0\times0.375\times(T - 3.8)=200.0\times4.184\times(85.0 - T)$$

Now, expand:

$$8.25(T - 3.8)=836.8(85.0 - T)$$

$$8.25T-31.35 = 71128-836.8T$$

$$8.25T + 836.8T=71128 + 31.35$$

$$845.05T=71159.35$$

Step4: Calculate T

$$T=\frac{71159.35}{845.05}\approx84.2^\circ C$$

Answer:

\(84.2\)