Question

7. A 0.70 kg cup [C 9x10J/kg. C ]is at 20° C. If 0.25 kg of water at 95° C is poured into the cup, what is the final temperature? 8. A 5.0x10 7 kg iron vessel and the 0.20 kg of water in it are at 20°C. If 8.0x10 * kg of metal shot at 100 C is dropped into it, the final temperature is 24°C. What is the specific heat of the metal? 9. A 0.21 kg brass calorimeter contains 0.10 kg of water. If 5x10J is required to raise the temperature of the water and container 10°C, what is the specific heat of brass? 10. If 0.20 kg of ice at-5°C is put into water at 0°C, how much water turns to ice?

Answer 1

Let the final temperature be T therefore Energy gained by cup = Energy least by water m_c C(T - 20) = m_w C_w (95 - T) C_w = 4.18 times 10^3 J/kg/degree C T - 20/95 - T = 0.25 times 4.18 times 10^3/0.7 times 9 times 10^2 T - 20/95 - T = 209/126 126 T - 20 times 126 = 209 times 95 - 209 T T = 20 times 126 + 209 times 95/126 + 209 = 66.79 degree C Heat lost by metal = Heat gained by Iron + Heat gained by water Specific heat of water C_w = 4.18 times 10^3 J/kg degree C Specific heat of Iron C_I = 4.5 times 10^2 J/kg degree C 8 times 10^-2 C_M (100 - 24) = 5 times 10^-2 times 450 times (24 - 20) + 0.2 times 4180 times (29 - 20) C_M = 5 times 10^-2 times 450 times 4 + 0.2 times 4180 times 4/76 times 8 times 10^-2 = 564.8 J/kg degree C