Question 1 A 0.0450 kg ice cube at -30.0°C is placed in 0.350 kg of 35.0°C water

Question

Question 1

A 0.0450 kg ice cube at -30.0°C is placed in 0.350 kg of 35.0°C water in a very well insulated container. What is the final temperature? degrees C

Question 2

An artist working on a piece of metal in his forging studio plunges the hot metal into oil in order to harden it. The metal piece has a mass of 54 kg and its specific heat is 0.1027 kcal/(kg · °C). He uses 810 kg of oil at 35°C. The specific heat of oil is 0.7167 kcal/(kg · °C). Once the metal is immersed in the oil, the temperature reaches an equilibrium value of 40°C. How hot was the forged metal piece just before he plunged it into the oil?
°C

Explanation / Answer

Q1 )

Given :-

mass of ice cube (Mi) = 0.0450 kg

Temp of ice (Ti) = -30.0°C

mass of water (Mw)= 0.350 kg

Temp of water (Tw) = 35.0°C

latent heat of fusion of water = 33.5x10^4 J/kg

latent heat of vaporization of water = 22.6x10^5 K/kg.

specific heat capacity of water = 4186 j/kg-C

Heat lost by water = Heat gain by ice cube = mc(DeltaT)

0.350 kg * ( 4186 j/kg-C ) (35 -T ) = 0.0450kg *  [33.5 * 10^4 J/kg + (4186 J/kgC)(T - 0 C) ]

=> 1465.1 (35 -T ) = 15263.37 (T - 0 C)

=>51278.5 - 1465.1 T = 15263.37 T -0

=> T = 3.35 °C ----------------------------------------------------(i) answer

B) Given :-

mass of metal (Mm) =54 kg

Specific Heat = 0.1027 kcal/(kg · °C)

Mass of oil (Mo) = 810 kg

Temp of Oil (To) = 35°C

specific heat of oil = 0.7167 kcal/(kg · °C)

Tequillibirum = 40°C

Again from formula,

Q = mc(Delta T)

Using,

Heat lost metal = Heat Gain by oil

54 * 0.1027 * (T - 40) = 810 * 0.7167 * (40 - 35)

5.5458 T - 221.832 = 2902.635

5.548 T = 3124.467

=> T = 563.169 °C -------------------------answer

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