Question

Constants Part A In a container of negligible mass, 0.270 kg of ice at an initial temperature of -28.0 °C is mixed with a mass m of water that has an initial temperature of 80.0° C. No heat is lost to the surroundings. If the final temperature of the system is 22.0 °C, what is the mass m of the water that was initially at 80.0° C? Express your answer with the appropriate units.

Answer 1

Mass of = 0.270 kg initial temperature = -28 degree c mass of water = m Initial temperature of water = 80.00 degree c now final temp of system = 22 degree c now at final heat gained by coaler mass = heat last by warmer mass now heat gained by = m_i times c_i times Delta theta [Because c of = 2100 J/kh k] = (0.270 kg) times (2100) times (22) = 12474 J rightarrow 1.247 times 10^4 J melting = (m_c times h_s) rightarrow (0.270 times 934 times 10^3 J/kg) [Because L of = 334 times 10^3] = 9.018 times 10^4 J melting water = (m_i times c_w times Delta theta) rightarrow (0.270 times 4190 times 22) [co f water = 4190 J/kg] = 2.488 times 10^4J Total heat gain = (1.247 + 9.018 + 2.488) times 10^4 J = 12.754 times 10^4 J now heat lass by water = (m_w times c_w times Delta theta) rightarrow m_w times 4190 times (80.28) = 2.430 times 10^5 m_w now Equating both 12.754 times 10^4 J times m_w m_w = 12.754