An ice cube with a mass

Question

An ice cube with a mass of 54.8 g at 0.0 °C is added to a glass containing 445 g of water at 45.0 °C. Determine the final temperature of the system at equilibrium. The specific heat capacity of water, C,, is 4.184 *C and the standard enthalpy of fusion, Delta H ba of water is 6.01 * 10 ^ 3 * j / m * Assume that no energy is transferred to or from the surroundings.

I promise I tried it and I didn't get it

Answer 1

According the the Law of Conservation of Energy, the heat lost by the water must be equal to the heat gained by the ice cube. We also know that the final temperature must be greater than oº and less than 45.0º. We will check our final answer to be sure this is the case.

We can write the heat lost by water as q = mC∆T

q = heat lost = ?

m = mass of water = 445 g

C = specific heat of water = 4.184 J/gº (you left out the units)

∆T = change in temperature = 45º - Tf where Tf is the final temperature that we are trying to find.

q = (445 g)(4.184 J/gº)(45 - Tf)

We can write the heat gained by the ice cube as q = m∆H_fusion + mC∆T

q = heat gained upon melting the ice cube (m∆Hf) + heat gained in raising the temperature from 0-Tf

m = mass of ice cube = 54.8 g

∆H_fusion = 6.01x10³ J/mol x 1 mol / 18 g = 334 J/g (change units to agree with others)

C = specific heat of water = 4.184 J/gº

∆T = change in temperature = Tf - 0

q = (54.8g)(334 J/g) + (54.8 g)(4.184 J/gº)( Tf - 0)

Now we can set these two equation equal to each other and solve for Tf (final temperature):

(445 g)(4.184 J/gº)(45 - Tf) = (54.8g)(334 J/g) + (54.8 g)(4.184 J/gº)( Tf - 0)

83785 - 1862Tf = 18303 + 229Tf - 0

2091Tf = 65482

Tf = 31.3ºC (note: final temperature is between 0º and 45º as expected).