Question

A 40-g block of ice is cooled to −68°C and is then added to 570 g of water in an 80-g copper calorimeter at a temperature of 28°C. Determine the final temperature of the system consisting of the ice, water, and calorimeter. (If not all the ice melts, determine how much ice is left.) Remember that the ice must first warm to 0°C, melt, and then continue warming as water. (The specific heat of ice is 0.500 cal/g · °C = 2,090 J/kg · °C.)

Answer 1

m_i = mass of the ice block = 40 g

T_ii = initial temperature of ice block = - 68 ^oC

T_f = final temperature of the system = ?

m_w = mass of water = 570 g

m_c = mass of copper calorimeter = 80 g

T_wi = initial temperature of water = 28 ^oC

T_ci = initial temperature of copper calorimeter = 28 ^oC

L = latent heat of fusion of ice to water = 80 cal/g

c_i = specific heat of ice = 0.5 cal/gC

c_w = specific heat of water = 1 cal/gC

c_c = specific heat of copper = 0.092 cal/gC

Using conservation of heat

heat gained = heat lost

m_i c_i (0 - T_ii ) + m_i L + m_i c_w (T_f - 0) = m_w c_w (T_wi - T_f ) + m_c c_c (T_ci - T_f )

(40) (0.5) (0 - (- 68)) + (40)(80) + (40) (1) (T_f - 0) = (570)(1)(28 - T_f ) + (80)(0.092)(28 - T_f )

T_f = 18.8 ^oC