How much energy is required to raise the temperature of 1.50 kg of cadmium from

Question

How much energy is required to raise the temperature of 1.50 kg of cadmium from 18.6°C to 168°C?
A 40 g block of ice is cooled to -70°C and is then added to 580 g of water in an 80 g copper calorimeter at a temperature of 25°C. Determine the final temperature of the system consisting of the ice, water, and calorimeter. Remember that the ice must first warm to 0°C, melt, and then continue warming as water. The specific heat of ice is 0.500 cal/g ·°C = 2090 J/kg°C. A 268g block of gold at 84°C is immersed in 155 g of water at 25.0°C. Find the equilibrium temperature, assuming the system is isolated and the heat capacity of the cup can be neglected How much energy is required to raise the temperature of 1.50 kg of cadmium from 18.6°C to 168°C?
A 40 g block of ice is cooled to -70°C and is then added to 580 g of water in an 80 g copper calorimeter at a temperature of 25°C. Determine the final temperature of the system consisting of the ice, water, and calorimeter. Remember that the ice must first warm to 0°C, melt, and then continue warming as water. The specific heat of ice is 0.500 cal/g ·°C = 2090 J/kg°C. A 268g block of gold at 84°C is immersed in 155 g of water at 25.0°C. Find the equilibrium temperature, assuming the system is isolated and the heat capacity of the cup can be neglected

Explanation / Answer

(a) Mass m = 1.5 Kg initial temperature t1 = 18.6°C Final temperature t2 = 168°C specific heat of cadmium s = 233 J/Kg°C Energy required Q = ms(t2 - t1)                            = 1.5*233*149.4                            = 52215.3 J (b) Mass of ice m1 = 40 g = 0.04 Kg mass of water m2 = 580 g= 0.58 Kg mass of copper m3 = 80 g = 0.08 Kg Initial temperature of ice t1 = -70°C Initial temperature of water & copper t2 = 25°C common temperature of ice,water & calorimeter t3 = ? specific heat of ice s1= 2090 J/kg°C specific heat of water s2 = 4186 J/kg°C specific heat of copper s3 = 390 J/kg°C Latent heat of fusion L = 333*10^3 J/kg According to the law of calorimetry Heat absorbed by ice during the change of phase + Heat absorbed by ice during the change of temperature = Heat given by water + Heat given by calorimeter m1L + m1s1(t3 - t1) = m1s2(t2 - t3) + m3s3(t2 - t3) m1L + m1s1(t3 - t1) = [m1s2 + m3s3](t2 - t3) (0.04*333000) + [(0.04*2090)(t3 - (-70))] = [(0.58*4186) + (0.08*390)](25 - t3) 13320 + 83.6(t3 + 70) = 2459.08(25 - t3) 2542.68t3 = 61477 - 13320 - 5852 2542.68t3 = 42305 t3 = 16.638°C So the final temperature of the system t3 = 16.638°C (c) Mass of gold block m1 = 268 g = 0.268 Kg mass of water m2 = 155 g = 0.155 Kg specific heat of gold s1 = 130J/kg°C specific heat of water s2 = 4186 J/kg°C initial temperature of gold t1 = 84°C initial temperature of water t2 = 25°C Let t3 be the common temperature. According to the law of calorimetry Heat lost by gold = Heat gained by water m1s1(t1 - t3) = m2s2(t3 - t2) 0.268*130*(84 - t3) = 0.155*4186*(t3 - 25) 34.84(84 - t3) = 648.83(t3 - 25) 2926.56 - 34.84t3 = 648.83t3 - 16220.75 19147.31 = 683.67t3 t3 = 28°C The common temperature t3 = 28°C (a) Mass m = 1.5 Kg initial temperature t1 = 18.6°C Final temperature t2 = 168°C specific heat of cadmium s = 233 J/Kg°C Energy required Q = ms(t2 - t1)                            = 1.5*233*149.4                            = 52215.3 J (b) Mass of ice m1 = 40 g = 0.04 Kg mass of water m2 = 580 g= 0.58 Kg mass of copper m3 = 80 g = 0.08 Kg Initial temperature of ice t1 = -70°C Initial temperature of water & copper t2 = 25°C common temperature of ice,water & calorimeter t3 = ? specific heat of ice s1= 2090 J/kg°C specific heat of water s2 = 4186 J/kg°C specific heat of copper s3 = 390 J/kg°C specific heat of copper s3 = 390 J/kg°C Latent heat of fusion L = 333*10^3 J/kg According to the law of calorimetry Heat absorbed by ice during the change of phase + Heat absorbed by ice during the change of temperature = Heat given by water + Heat given by calorimeter m1L + m1s1(t3 - t1) = m1s2(t2 - t3) + m3s3(t2 - t3) m1L + m1s1(t3 - t1) = [m1s2 + m3s3](t2 - t3) m1L + m1s1(t3 - t1) = [m1s2 + m3s3](t2 - t3) (0.04*333000) + [(0.04*2090)(t3 - (-70))] = [(0.58*4186) + (0.08*390)](25 - t3) 13320 + 83.6(t3 + 70) = 2459.08(25 - t3) 2542.68t3 = 61477 - 13320 - 5852 2542.68t3 = 42305 t3 = 16.638°C So the final temperature of the system t3 = 16.638°C (c) Mass of gold block m1 = 268 g = 0.268 Kg mass of water m2 = 155 g = 0.155 Kg specific heat of gold s1 = 130J/kg°C specific heat of water s2 = 4186 J/kg°C initial temperature of gold t1 = 84°C initial temperature of water t2 = 25°C Let t3 be the common temperature. According to the law of calorimetry Heat lost by gold = Heat gained by water m1s1(t1 - t3) = m2s2(t3 - t2) 0.268*130*(84 - t3) = 0.155*4186*(t3 - 25) 34.84(84 - t3) = 648.83(t3 - 25) 2926.56 - 34.84t3 = 648.83t3 - 16220.75 19147.31 = 683.67t3 t3 = 28°C The common temperature t3 = 28°C

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.