Question

Q 4. A 0.5 kg of aluminum cube at 27 ℃ and a 1.0 kg of copper cube at 100 ℃ are placed 

in 2 kg of water at 22 ℃. Assuming no heat loss to the surroundings, calculate the final 

temperature of the mixture.

Specific heat capacities: water 𝑐𝑊 = 4187 𝐽𝑘𝑔−1𝐾

−1

, copper 𝑐𝑐𝑢 = 386 𝐽𝑘𝑔−1𝐾

−1

, 

aluminum 𝑐𝑎𝑙 = 900 𝐽𝑘𝑔−1𝐾

−1

Q 5. A 5 kg of ice cube at −10℃ is mixed with 0.1 kg of water at 80 ℃. There is no heat 

loss to the surrounding. The specific heat capacity of ice is 2.22 𝑘𝐽𝐾

−1𝑘𝑔−1

,The specific 

heat capacity of water is 4.187 𝑘𝐽𝐾

−1𝑘𝑔−1

. The specific latent heat of fusion of ice is 

333 𝑘𝐽 𝑘𝑔−1

.

(a) What is the final physical state of the mixture? (Gas, Liquid or Solid?)

Justify your answer using suitable calculations 

(b) What is the final temperature of the mixture?

Answer 1

Q4. Let the final temperature o the mixture =$x^0C$

Heat lost by copper = heat gained by aluminium and water

$m_cc_c(100-x)=m_wc_w(x-22)+m_{al}c_{al}(x-27)$

$1\times 386(100-x)=2\times 4187(x-22)+0.5\times 900(x-27)$

$38600-386x=8374x-184228+450x-2150$

$38600+184228+2150=386x+8374x+450x$

$234978=9210x$

$x=25^0C$

Q5. (a) Heat lost by water = heat gained by ice at -10$^0C$ +heat of fusion of ice +heat of melted ice (water)

Let the final temperature =$x^0C$

$m_wc_w(80-x)=m_{ice}c_{ice}(x+10)+m_{ice}L_f+m_{w}c_w(x-0)$

$0.1\times 4187(80-x)=5\times 2220(x+10)+5\times 333000+5\times 4187(x)$

$33496-418.7x=11100x+111000+1665000+20935x$

$-32453.7x=1742504$

$x=-53.69^0C$

Final temperature is negative, so the mixture is a solid.

b) Final temperature is $53.69^0C$