Question

Solid ice (25 g) was added to a glass of water (400 g) which was initially at 25ºC. The final equilibrium temperature that the glass of water will be is TºC. (Specific heat capacity of water = 4.19 J/g∙ºC and specific latent heat of fusion = 334 J/g) (a) Express in terms of T for the heat energy lost by the water to reach equilibrium temperature of TºC? (b) Calculate the heat energy gained by the ice for complete melting of ice without any change in temperature? (c) What will be the heat energy gained by the ice after complete melting to reach equilibrium temperature of TºC? (d) Since the heat energy lost by the water is equal to the sum of heat energy gained by the ice, calculate the equilibrium temperature reached for the glass of water.

Answer 1

Given

Mass of the ice , m_ice = 25 g

Mass of the glass of water , m_w = 400 g

Initial temperature,T₁  = 25^Oc

Final equilibrium temperature that the glass of water,T₂  = T ^Oc

Specific heat capacity of water, c_p= 4.19 J/g^Oc

specific latent heat of fusion,L = 334 J/g

-----------------

a)

Find the heat energy lost by the water to reach equilibrium temperature of TºC:

Q1 = m_wc_p ∆T

       = (400)(4.19)( T - 25 )

     =1676( T - 25 ) J

b)

Find the heat energy gained by the ice for complete melting of ice without any change in temperature:

Q2 =  m_iceL

      = (25)(334)

     = 8350 J

-----------------------------

c)

Find the heat energy gained by the the ice after complete melting to reach equilibrium temperature of TºC :

Q3 = m_icec_p ∆T

       = (25)(4.19)( T - 25 )

     =104.75( T - 25 ) J

d)

Calculate the equilibrium temperature reached for the glass of water by using the conservation energy:

Q1 =Q2+Q3

1676( T - 25 )=8350+104.75( T - 25 )

1676T-41900=8350+104.75T-2618.75

1676T-104.75T=8350+41900-2618.75

1571.25T=47631.25

T= 30.31^Oc