[samiam]

6.66. Use the 334 J/g (0.334 kJ/g) heat of fusion to calculate the heat of melting. Then use the heat of
melting to calculate Δt, the temperature change.
Heat of melting = x 31.5 g x 334 J
1 g
= 10521 J
Heat melting + heat warming = heat lost by 210 g H2O
Let t = the final temperature, and substitute into the above equation:
10521 J + 4.18 J1/ g • C x 31.5 g x (t − 0°C) = 4.18 J/1 g • Cx 210 g x (21.0°C − t)
Solve the above equation for t.
10521 J + 131.67t = 18433.8 J − 877.8t
t = 18433.8 J - 10521 J
(131.67 + 877. J/°C
= 7.838 = 7.84°

for the +heat warming---> why is tfinal 0?

[opti384]

What do you mean by "for the +heat warming---> why is tfinal 0?"

Are you referring to this part 4.18 J1/ g • C x 31.5 g x (t − 0°C) asking why it's t − 0°C?

[samiam]

What do you mean by "for the +heat warming---> why is tfinal 0?"

Are you referring to this part 4.18 J1/ g • C x 31.5 g x (t − 0°C) asking why it's t − 0°C?

yeah

[samiam]

Oops I totally forgot to write the question out..
When ice at o degrees Celsius melts to liquid water 0 degrees Celsius
it absorbs 0.334kj of heat per gram. suppose the heat needed to melt 31.5g of ice is absorbed from the water contained in a glass. If this water has a mass of 0.210 kg and a temp off 21 degrees celsius what is the final temp of the water(note you will also gt 31.5g of water at  0 degrees celsius)

6.66. Use the 334 J/g (0.334 kJ/g) heat of fusion to calculate the heat of melting. Then use the heat of
melting to calculate Δt, the temperature change.
Heat of melting = x 31.5 g x 334 J
1 g
= 10521 J
Heat melting + heat warming = heat lost by 210 g H2O
Let t = the final temperature, and substitute into the above equation:
10521 J + 4.18 J1/ g • C x 31.5 g x (t − 0°C) = 4.18 J/1 g • Cx 210 g x (21.0°C − t)
Solve the above equation for t.
10521 J + 131.67t = 18433.8 J − 877.8t
t = 18433.8 J - 10521 J
(131.67 + 877. J/°C
= 7.838 = 7.84°

for the +heat warming---> why is tfinal 0?

[opti384]

What do you mean by "for the +heat warming---> why is tfinal 0?"

Are you referring to this part 4.18 J1/ g • C x 31.5 g x (t − 0°C) asking why it's t − 0°C?

Well as the ice melts at 0°C it absorbs the heat of fusion and becomes water which has the temperature of 0°C.

Now the heat from the water in the glass is transferred to the water (0°C) from the ice. This 4.18 J1/ g • C x 31.5 g x (t − 0°C) is used to find that heat.