Question

a 44.0 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 250.0 g of water at 30.0 °c and a constant pressure of 1 atm. the initial temperature of the brass is 81.3 °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 3 significant digits. ☐ °c

Explanation:

Step1: Define heat transfer equation

Heat lost by brass = heat gained by water: $m_b c_b (T_{b,i} - T_e) = m_w c_w (T_e - T_{w,i})$
where $m_b=44.0g, c_b=0.375Jg^{-1}^\circ C^{-1}, T_{b,i}=81.3^\circ C$; $m_w=250.0g, c_w=4.184Jg^{-1}^\circ C^{-1}, T_{w,i}=30.0^\circ C$; $T_e$=equilibrium temp.

Step2: Substitute values into equation

$44.0 \times 0.375 \times (81.3 - T_e) = 250.0 \times 4.184 \times (T_e - 30.0)$

Step3: Simplify both sides

Left: $16.5 \times (81.3 - T_e) = 1341.45 - 16.5T_e$
Right: $1046 \times (T_e - 30.0) = 1046T_e - 31380$

Step4: Solve for $T_e$

$1341.45 - 16.5T_e = 1046T_e - 31380$
$1341.45 + 31380 = 1046T_e + 16.5T_e$
$32721.45 = 1062.5T_e$
$T_e = \frac{32721.45}{1062.5} \approx 30.8^\circ C$

Answer:

30.8