Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°c and 237 g of water initially at 100.0°c.

115 ⁰C

Explanation:

Step 1: The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

-----eqution 1

where,

is the heat absorbed by the solid at 0⁰C

is the heat absorbed by the liquid at 0⁰C

the heat lost by the warmer water sample

Important equations to be used in solving this problem

, where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

is change in temperature

Again,

-------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

Step 2: calculate how many moles of water you have in the 100.0-g sample

Step 3: calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

This means that equation (1) becomes

79.13 KJ +

Step 4: calculate the final temperature of the water

Substitute in the values; we will have,

79.13 kJ + 990.66J* = -1463J*

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* = -1.463KJ*

collect like terms,

2.45366 = 283.133

∴ = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

the minimum amperage that the current in this wire would have to have so that its weight is completely supported by the magnetic force due to earth's field, assuming that no other forces except gravity act on it