Question

A = 13 B = 27

A (10.0+A) g ice cube at -15.0^oC is placed in (125+B) g of water at 48.0^oC. Find the final temperature of the system when equilibrium is reached. Ignore the heat capacity of the container and assume this is in a calorimeter, i.e. the system is thermally insulated from the surroundings. Give your answer in^oC with 3 significant figures.

Specific heat of ice: 2.090 J/g K Specific heat of water: 4.186 J/g K Latent heat of fusion for water: 333 J/g

Answer 1

mass of ice= 10+13 =23g , mass of water = 125+27= 152g

energy lost by water = energy gained by ice

or, mass of water*heat capacity of water*temperature change for water = mass of ice*heat capacity of ice*temperature change for ice + latent heat of fusion*mass of ice +mass of ice converted to water*heat capacity of ice converted to water*temperature change for ice converted to water

So, 152*4.186*(48-T) = 23*2.09*(0-(-15)) + 333*23 + 23*4.186*(T-0)

So, 152*4.186*(48-T) = 23*2.09*15 + 333*23 + 23*4.186*T

So, -636.272T - 96.278T = 8380.05 - 30541

So, T= 22160.95 / 732.55

So, the final temperature of the system when equilibrium is reached is T = 30.3 oC