Question

A 40-g block of ice is cooled to −72°C and is then added to 590 g of water in an 80-g copper calorimeter at a temperature of 26°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

Let the resulting temperature be

Given mass of water , m₁ = 590 g , specific heat of water = S₁ = 1 cal / g/ , mass of calorimeter w = 80 g

mass of ice = m₂ = 40 g, specific heat of ice = S₂ = 0.5 cal / g/

Let be the change in temperature & L be the latent heat , for ice L = 80 cal /g

Heat lost by water =( m₁ + w ) S₁ = cal

Heat gained by ice

i) from to = m₂ S₂ =

ii) in converting into water at = m₂ L = 40 80 = 3200 cal

iii) in raising the temperature of water formed from to = m₂ S₁

According to principle of calorimetry, heat lost = heat gained

Heat lost by warm water = =

Amount of ice melted by 2991.3 cal =

where latent heat of fusion of water = 335000 J / kg = 335 J / g = 80.0670cal /g

Remaining quantity of ice = 40 - 37.36 = 2.64 g