Question

please help

Answer 1

@e.mccormick @iambatman @iGreen @hartnn

Answer 2

a heat balanced. heat lost by hot = heat gained by cold substance
2) heat = Q = m Cp dT for a temp change without a phase change
3) heat = Q = m x dHfus, vap, sub, etc. for a phase change without a temp change.

*********
your problem...

the solid heats up from 24 to 27 C.. then melts at 27 C. then the liquid warms to a max of 700C. then vaporizes. then the gas heats up to whatever. ok?

Q = (m Cp dT) solid + (m dHfus) + (m Cp dT)liquid + (m dHvap) + (m Cp dT)vapor.

do you see that? so let's take each of those terms one at a time...
(m Cp dT) solid = 50g x 0.48cal/g°C x (27-24 °C) = 72 cal

you're adding more heat than that... right? 5 kcal = 5000 cal ? so let's continue.
(m dHfus) = 50g x 45cal/g = 2250cal..

now you this much heat left... 5000 cal - 72 cal - 2250 cal = 2678 cal.. let's continue...

IF we heated all the resulting liquid to it's boiling point we would need this much heat...
Q = m Cp dT = 50g x (0.75cal/g°C) x (700 - 27 °C) = 25,238 cal.

we don't have that much available. we only have 2678 available.. so we can't heat it to boiling.. if fact we have this...
2678 cal = m Cp dT
2678 cal = 50g x (0.75 cal/g°C) x (Tf - 27°C)
71.4°C = (Tf - 27°C)
Tf = 98°C

so the final temp is 98°C and the substance is in liquid phase.

Answer 3

Might onet to ask in chem. This is one of those ones where I think you need to understand the chem part to do the math. Someone like @abb0t could explain the chem to you... but I do not see him on.

Answer 4

Thanks @goodday443

Answer 5

a heat balanced. heat lost by hot = heat gained by cold substance 2) heat = Q = m Cp dT for a temp change without a phase change 3) heat = Q = m x dHfus, vap, sub, etc. for a phase change without a temp change. ********* your problem... the solid heats up from 24 to 27 C.. then melts at 27 C. then the liquid warms to a max of 700C. then vaporizes. then the gas heats up to whatever. ok? Q = (m Cp dT) solid + (m dHfus) + (m Cp dT)liquid + (m dHvap) + (m Cp dT)vapor. do you see that? so let's take each of those terms one at a time... (m Cp dT) solid = 50g x 0.48cal/g°C x (27-24 °C) = 72 cal you're adding more heat than that... right? 5 kcal = 5000 cal ? so let's continue. (m dHfus) = 50g x 45cal/g = 2250cal.. now you this much heat left... 5000 cal - 72 cal - 2250 cal = 2678 cal.. let's continue... IF we heated all the resulting liquid to it's boiling point we would need this much heat... Q = m Cp dT = 50g x (0.75cal/g°C) x (700 - 27 °C) = 25,238 cal. we don't have that much available. we only have 2678 available.. so we can't heat it to boiling.. if fact we have this... 2678 cal = m Cp dT 2678 cal = 50g x (0.75 cal/g°C) x (Tf - 27°C) 71.4°C = (Tf - 27°C) Tf = 98°C so the final temp is 98°C and the substance is in liquid phase.

3 minutes ago

99