
College Physics
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ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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How much energy is required to increase the temperature of 25 g water from 0 C∘ to 50 C∘. The specific heat capacity of water is 1 cal/g⋅c∘
(Hint: Specific heat Q and 1 cal = 4.2J)
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- A 20-g ice cube floats in 180 g of water in a 100-g copper cup; all are at a temperature of 0°C. A piece of lead at 90°C is dropped into the cup, and the final equilibrium temperature is 12°C. What is the mass of the lead? (The heat of fusion and specific heat of water are 3.33 105 J/kg and 4,186 J/kg · °C, respectively. The specific heat of lead and copper are 128 and 387 J/kg · °C, respectively.)arrow_forwardIf 31.0 g of milk at a temperature of 4.05°C is added to a 260-g cup of coffee at a temperature 91.0°C, what is the final temperature of the mixture? Assume coffee has a specific heat of 4.19 ✕ 103 J/(kg · K) and milk has a specific heat of 3.93 ✕ 103 J/(kg · K).arrow_forwardSuppose you want to raise the temperature of a 0.185-kg piece of ice from -20.0°C to 130°C. The heat of fusion is Lf = 334 kJ/kg, and the heat of vaporization is Lv = 2256 kJ/kg. In this problem, take 1520 J/kg⋅°C as the specific heat of steam, and 2090 J/kg⋅°C as the specific heat of ice. Part (a) How much heat, in kilocalories, must be transferred for this to happen, including the energy needed for phase changes? Part (b) How much time, in seconds, is required to do this, assuming a constant 20.0 kJ/s rate of heat transfer?arrow_forward
- When a material changes phase - liquid, solid, gas- heat exchange is required. This heat is called the Latent heat. is the equation where L is the Latent heat and is material specific. How much heat is required to take water (mass of .55 kg) from -20 to 20 C? L for ice to water is 33.5 * 104 J/kg. the specific heat of ice is 2090 J/(kg*C).and for water is 4186 J/(kg*C).arrow_forwardIce at 0 °C is placed in a Styrofoam cup containing 0.62 kg of lemonade at 32 °C. The specific heat capacity of lemonade is virtually the same as that of water; that is, c = 4180 J/(kg C°). After the ice and lemonade reach an equilibrium temperature, some ice still remains. The latent heat of fusion for water is Lf = 3.35 x 105 J/ kg. Assume that the mass of the cup is so small that it absorbs a negligible amount of heat, and ignore any heat lost to the surroundings. Determine the mass of ice that has melted, in grams.arrow_forwardWhat is the ratio of the energy required to warm 125 g of Ice (0.0 \deg C) to body temperature (37 \deg C) to warming the same amount of water through the same temperature change? Answer format is the number Eice/Ewater = (2 significant figures) Latent Heat of Fusion of Water: 335, 000 J/kg Specific Heat Capacity of Water: 4186 J/kg/\deg Carrow_forward
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