An ice cube of mass 9.0 g at temperature 0∘c is added to a cup of coffee, whose temperature is 90 ∘c and which contains 110 g of liquid. assume the specific heat capacity of the coffee is the same as that of water. the heat of fusion of ice (the heat associated with ice melting) is 6.0 kj/mol.

Answer : The temperature of the coffee after the ice melts is

Explanation :

In this question we have to determine the temperature of the coffee after the ice melts.

Heat lost by coffee = Heat gained by ice

where,

= specific heat of coffee = specific heat of water =  (as per question)

= specific heat of ice =

= mass of coffee = 110 g

= mass of ice = 9.0 g

= final temperature = ?

= initial temperature of coffee =

= initial temperature of ice =

n = number of moles of ice =

= 6.0 kJ/mol = 6000 J/mol

Now put all the given values in the above formula, we get

Therefore, the temperature of the coffee after the ice melts is

altitude, density, pressure

air gets less dense the higher the altitude and also the air pressure is less as altitude rises.

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6