Question

1. In a light foam cooler is a 2.4 kg slab of aluminum at 20C. I add to it 300 g of ice at -10C. What is the equilibrium temperature? How much of the ice melts?

2. If I add 30 g of ice at -15C to 100 g of water at 75C, what is the equilibrium temperature?

3. If 12 g of steam at 100C condenses on a 1500-g iron bar initially at 10C, what will the equilibrium temperature be?

Answer 1

1) given :

M = mass of aluminium = 2.4 kg

intial temperature of aluminium = T1 = 20 degree celsius

m = mass of ice added = 300 g = 0.3 kg

intial temperature of ice = T2 = -10 degree celsius

specific heat of alumium = S = 921 J/kg/K

specific heat of ice = s = 2108 J/kg/K

latent heat of fusion = L = 333000 J/kg

Lets assume the final temperature be 0.

then heat lost by Aluminium = Q1 = M*S*(T1-0) = 2.4*921*20 = 44208 J

Heat gain by ice when moving from -10 celsius to zero celsius = Q2 = m*s*(0-T2) = 0.3*2108*10 = 6324J

Heat gain by ice to change phase at zero celsius = Q3 = mL = 0.3*333000 = 99900 J

As Q2 +Q3 is greater than Q1 but Q2 is less than Q1, it means total ice reached 0 degree celsius but did not converted into water. Therefore, final temperature is zero degree celsius. [answer]

2) given :

M = mass of ice = 30 g

T1 = initial temperature of ice = -15 celsius

m = mass of water = 100 g

T2 = initial temperature = 75 celsius

S = specific heat of ice = 2.1 J/g/K

s = specific heat of water = 4.2 J/g/K

L = latent heat of fusion = 333 J/g/K

Lets assume the final temperature be T

Heat lost by water = Q1 = ms(T2-T)

Heat gained by ice from -15 to zero = Q2 = MS(0-T1)

Heat gain by ice when converting into water = Q3 = ML

Heat gain by this water from zero to T = Q4 = Ms(T-0)

Using Principle of Calorimetry :

Q1 = Q2 + Q3 + Q4

=> ms(T2-T) = MS(0-T1) + ML + Ms(T-0)

on solving we get, T = 37.66 degree celsius. [answer]