Question

Steam at 100°C is condensed into a 54.0 g copper calorimeter cup containing 280 g of water at 25.0°C. Determine the amount of steam (in g) needed for the system to reach a final temperature of 64.0°C. The specific heat of copper is 387 J/(kg·°C). 4231.29 Be sure to account for the heat energy absorbed by the calorimeter cup and the water in the cup, and the heat energy contributed by the steam. Note that the steam contributes heat energy by two processes: first, the condensation of steam into water, and second, the cooling of the associated water to the final temperature. The algebra involved is straightforward; however, since the problem has a large number of values, it is critical that you check your calculation carefully. g

Answer 1

Loss of energy = gain of the energy mL + mc_w delta T = m-1 c_w delta T + m_v C_cm delta T 1 = > mL + m c_v (100 - T) = m_1 c_w (T - 25 degree) + M_2 cos (T - 25 degree) 2 Here m = 2 T = 64 degree C m_1 = 280 g m_2 = 54 g Cu = 387 J/kg C_m = 0.387 J/g degree C, C_w = 4.186 J/g degree C L_v = 2.26 times 10^6 J/kg = 2.26 times 10^3 J/g Put these value in eqn 1 2.26 times 10^3 m + m (4.186) (100 - 64) = 280 times 4.186 (64.25) + 54 (0.387) (64 - 25) 2410.696 m = 54711.12 + 815.022 m = 46526.34/2410.696 m = 19.3 g