Answer:
Latent heat fusion of water, hf = 334 kJ/kg
specific heat of water is 4.186 kJ·kgâ’1·Kâ’1
specific heat of ice is 2.05 kJ/kgK
First, lets assume the ice will melt and become liquid water
Q in = Q out
m1*C(ice)*(0 deg C- (-) 15) + m1*hf + m1*C(water) (Tf - 0))= m2*C(water)*
(T2 - Tf)
1.5* 2.05*15 + 1.5*334 + 1.5*4.186*Tf = 0.1*4.186 (20 - Tf)
46.125 + 501 + 6.279 Tf = 0.4186 (20 - Tf)... (1)
For water if it decrease to 0 deg C, the energy as heat release is 0.4186 (20
- 0) = 8.372 kJ < 46.125 kJ, Lower than Ice which increase temperature from
-10 to zero. So Water will become ice.
The energy dissipate for 20 deg C water and change its phase to 0 deg C ice
is still lower than energy absorb by ice to 0 deg C, rearrange the equation.
Q in = Q out
m1*C(ice)*(Tf- (-) 15) = m2*C(water)*(T2 - 0) + m2*hf + m2*c(ice) (0-Tf)
1.5* 2.05* (Tf +15) = 0.4186*20 + 0.1*334 + 0.1*2.05 (-Tf)
3.075 (Tf + 15) = 8.372 + 33.4 + 0.205 (-Tf)
3.075Tf + 46.125 = 41.772 - 0.205 Tf
3.28 Tf = - 4.353
Tf = - 1.327 deg C ( Ice phase)
Well, even though it may look like the last option is the correct one, that one is wrong. When a crystal s formed from many metals there are many milecular orbitals. So the answer is B. Hope this is what you are looking for
The energy required to heat 40g of water from -7 c to 108 c is
1541000 joules
calculation
Q(heat)= M( mass) x c(specific heat capacity) xdelta t( change in temperature)
M= 40g= 40/1000= 0.04 Kg
C= 335,000 j/kg/c
delta T ( 108 --7= 115 c)
Q is therefore = 0.04 g x 335000 j/kg/c x 115 c = 1541,000 joules
