Question

A 170 g copper bowl contains 245 g of water, both at 20.0°C. A very hot 300 g copper cylinder is dropped into the water, causing the water to boil, with 5.45 g being converted to steam. The final temperature of the system is 100°C. Neglect energy transfers with the environment.

(c) What is the original temperature of the cylinder? (in celcius)

Answer 1

m_b = mass of bowl = 170 g

m_c = mass of copper cylinder = 300 g

m₁ = mass of water converted to steam = 5.45 g

m₂ = mass of water at temperature 100 = 239.55

T = original temperatrure of cylinder

c_c = specific heat of copper = 0.385

c_w = specific heat of water = 4.2

L = 2260 J/g

using conservation of energy

m_c c_c (T - 100) = m_b c_c (100 - 20) + m₂ c_w (100 - 20) + m₁ c_w (100 - 20) + m₁ L

(300) (0.385) (T - 100) = (170) (0.385) (80) + (239.55) (4.2) (80) + (5.45) (4.2) (80) + (5.45) (2260)

T = 964.7