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How many joules of heat are needed to melt 50 g of ice at 0°C and then warm the liquid to 65°C? heat of fusion of ice = 334 J/g specific heat of water = 4.184 J/g°C specific heat of ice = 2.03 J/g°C heat of vaporization of water = 2260J/g

User Bjorne
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2 Answers

2 votes

Answer:

30,298 joules of heat are needed to melt 50 g of ice at 0°C and then warm the liquid to 65°C

Step-by-step explanation:

To analyze the heat that must be supplied to the ice, two phases are analyzed: one in which a part of heat will be required to melt the ice, that is, convert 0 ° C of ice to 0 ° C in a liquid state, and another part of heat that will raise the temperature of the melted ice to 65 ° C. So:

Total heat required = Heat required to melt ice + Heat required to raise the temperature of the ice in liquid state

Being fusion, the process that a substance undergoes to go from a solid state to a liquid, then the heat required to convert 0 ° C of ice to 0 ° C in a liquid state is calculated as:

Heat required to melt ice=mass*heat of fusion of ice= 50 g* 334
(J)/(g)= 16,700 J

On the other hand, the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous) is calculated by the expression:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation (Tfinal-Tinitial).

Then, the heat required to go from water at 0 ° C to water at 65 ° C is calculated by:

q=specific heat of water*m*ΔT= 4.184
(J)/(g*C) *50 g* (65 °C - 0°C)= 13,598 J

So:

Total heat required = 16,700 J + 13,598 J

Total heat required = 30,298 J

30,298 joules of heat are needed to melt 50 g of ice at 0°C and then warm the liquid to 65°C

User Nicolas Buquet
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5.8k points
1 vote

Answer:


Q=30298J

Step-by-step explanation:

Hello,

In this case, in this heating process, we should consider two steps which have their own associated enthalpy for the same mass of water:

1) Melting of ice at 0 °C.


H_1=m_(ice)\Delta _(fusion)H=50g*334(J)/(g) =16700J

2) Heating of liquid water from 0 °C to 65 °C.


H_2=mCp(T_2-T_1)=50g*4.184(J)/(g\°C)(65-0)\°C =13598J

Therefore, the total needed heat turns out:


Q=H_1+H_2=16700J+13598J\\\\Q=30298J

Regards.

User ChristianFigueroa
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