Question

Please Show all work, thank you!

A copper block with a mass of 400 grams is cooled to 77 K by being immersed in liquid nitrogen. The block is then placed in a Styrofoam cup containing some water that is initially at +50.0°C. Assume no heat is transferred to the cup or the surroundings. The specific heat of liquid water is 4186 J/(kg °C), of solid water is 2060 J/(kg °C), and of copper is 385 J/(kg °C). The latent heat of fusion of water is 3.35 105 J/kg. What is the mass of water in the cup, if the final temperature is −25.0°C?

_______ g

Answer 1

Mass of copper block = m_c = 400 g = 0.4 kg

Initial temperature of copper = T_c = 77 K = -196 ^oC

Mass of water = m_w

Initial temperature of water = T_w = 50 ^oC

Final temperature of mixture = T = -25 ^oC

Freezing temperature of water = T_f = 0 ^oC

Specific heat of liquid water = C_w = 4186 J/(kg ^oC)

Specific heat of solid water = C_i = 2060 J/(kg ^oC)

Specific heat of copper = C_c = 385 J/(kg ^oC)

Latent heat of fusion of water = L = 3.35 x 10⁵ J/kg

The heat lost by water is the heat gained by copper.

m_cC_c(T - T_c) = m_wC_w(T_w - T_f) + Lm_w + m_wC_i(T_f - T)

(0.4)(385)(-25 - (-196)) = m_w[4186(50 - 0) + (3.35x10⁵) + 2060(0 - (-25)]

26334 = m_w(209300 + 335000 + 51500)

m_w = 0.0442 kg

m_w = 44.2 grams

Mass of water in the cup = 44.2 grams