a)You are a blacksmith and have been working with 19.3 kg of steel. When you are
ID: 1727876 • Letter: A
Question
a)You are a blacksmith and have been working with 19.3 kg of steel. When you are finished shaping it,the steel is at a temperature of 231°C. To cool it off, you drop it into a bucketcontaining 3.7 kg of water at 68°C. How much of this water is converted tosteam? Assume the steel, the water, and the steam all have the samefinal temperaturek.b)The author's gold wedding ring is stuck on his finger, andhe must remove it. The ring has an inside diameter of 1.96 cm, but the knuckle of the finger in question hasa diameter of 2.135 cm. The author decidesto get the ring off by heating it. How much must the temperature ofthe ring be increased for it to slide past the knuckle? Assume theauthor's finger does not change size when it is heated.
a)You are a blacksmith and have been working with 19.3 kg of steel. When you are finished shaping it,the steel is at a temperature of 231°C. To cool it off, you drop it into a bucketcontaining 3.7 kg of water at 68°C. How much of this water is converted tosteam? Assume the steel, the water, and the steam all have the samefinal temperaturek.
b)The author's gold wedding ring is stuck on his finger, andhe must remove it. The ring has an inside diameter of 1.96 cm, but the knuckle of the finger in question hasa diameter of 2.135 cm. The author decidesto get the ring off by heating it. How much must the temperature ofthe ring be increased for it to slide past the knuckle? Assume theauthor's finger does not change size when it is heated.
Explanation / Answer
(a) Mass of steel, M1 = 19.3 kg Initial temperature of steel, t1 = 231oC Mass of water, M2 = 3.7 kg Initial temperature of water, t2 = 68oC Specific of steel, C1 = 500 J / kg .oC Specific heat of water , C2 = 4186 J / kg .oC Temperature of steam, t3 = 100 oC Latent heat of steam, L = 2260440 J / kg Heat lost by steel during the fall of temp from231 oC to 100 oC = M1 C1T = 19.3 * 500 * 131 = 1264150 J Heat gained by water to during the raise of tempfrom 68 oC to 100 oC = M2 C2 T = 3.7 * 4186 * 32 = 495622.4 J Let Mass of the water converted into steam = x kg Heat energy required by x kg of water to convertinto steam = m L = x * 2260440 J Hence, x * 2260440 = 1264150 - 495622.4 Mass of the water converted into steam, x = 0.34kg (b) Initial diameter, L = 1.96 cm Final diameter, L' = 2.135 cm Coefficient of linear expansion of gold, = 14.1 X 10 -6 / oC Raise in temperature, T = ? Increase in diameter, L = 0.175 cm Increase in diameter, L = LT Raise in temperature, T = L / ( L ) = 0.175 / ( 14.1 X 10 -6 * 1.96 ) = 6332.3 oc
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