What is the temperature of water at room temperature?
First, let’s just answer the simpler question: “What is room temperature”?Room temperature is commonly given as 20 degrees Celsius (which is 68 degrees Fahrenheit). However, it may also be given as a narrow range of temperature, like 20 to 22 degrees Celsius (68 to 72 degrees Fahrenheit). (NOTE: 71.6 degrees got rounded up to 72 degrees.)If the air in the room goes through temperature variations, then a body of water (contained in a sealed water bottle) will heat up or cool down with the air but will lag behind it because it has “heat capacity” and a limited rate at which heat transfers into or out of the bottle of water. If the air temperature in the room fluctuates back and forth very quickly between two extremes, then the temperature of the water in the bottle will tend to average out those temperature variations, so its temperature extremes will be much closer together (very close to the average ambient temperature). For example, the air in the room might vary between 20 and 22 degrees C, but the water in the bottle might vary between 20.8 degrees C and 21.2 degrees C, depending on how quickly the ambient air temperature cycles back and forth.If the water is poured onto a piece of highly absorbent cloth (like a towel) and is exposed to “room temperature” air that is extremely dry and has a fan blowing on it, then it will evaporate quickly. The heat absorbed by the rapid evaporation will cool down the damp towel. The temperature of the damp towel could easily dip down to something like 10 degrees C, turning it into what is commonly known as a “swamp cooler” — until all the water evaporates and the towel becomes dry. Or, until the humidity level in the room doesn’t allow any further evaporation (an equilibrium gets reached between evaporation and condensation).
A cup of coffee at 170 degrees is poured into a mug and left in a room at 65 degrees. After 1 minutes, the cof?
Newton's law of cooling states that: dT(t)/dt = -k*(T(t) - Ts) where k is a positive constant, T is the temperature at time t, and Ts is the temperature of the surroundings. We can separate this differential equation to get: dT/(T - Ts) = -k dt If Ts is assumed to be constant, we can integrate to get: ln(T - Ts) - ln(c) = -k*t where ln(c) is the constant of integration. If the temperature at t = 0 is To, then: ln((To-Ts) - ln(c) = 0 c = (To - Ts) So: ln((T - Ts)/(To - Ts)) = -k*t At this point, we can use the fact that at t = 1 min, the coffee has cooled from To = 170 to T(1min) = 130 in order to solve for the time constant "k": ln((130 - 65)/(170 - 65)) = -k*(1min) ln(65/105) = -0.48 = -k*min k = 0.48/min Now solve the above equation for T: T(t) = Ts + (To - Ts)*exp(-k*t) Plugging in the values of the constants for this question: T(t) = [65 + 105*exp(-0.48*t/min)]*degrees Plug in t = 11min to find that after that amount of time, the coffee will be 65.5 degrees. TO solve for the time at which the coffee will be a certain temperature, it's easier to go back to the equation: ln((T - Ts)/(To - Ts)) = -k*t Solving for when the coffee is 100 degrees: -(1min/0.48*ln((100 - 65)/(170 - 65)) = t(T) t(100 deg) = 2.29 min
Temperature and specific heat problem?
Heat Lost by coffee= Heat gained by spoon Relationship: Q = m* c * dT, where m is the mass, c is the specific heat capacity and dT is the "change" in temperature. Assume the final temperature of the whole system is "x". Heat lost by coffee = 180 * 4.184 * (95 - x) (95 -x) because coffee will loose heat, hence temperature of coffee will fall. Heat gained by spoon = 45 * 0.24 * (x - 25). Heat lost = Heat gained : 180 * 4.184 * (95 - x) = 45 * 0.24 * (x - 25), Solve for x which is the FINAL temperature of the system. So (95 -x) will be the temperature reduced for coffee (ANSWER)
A piece of copper metal is initially at 100º C. It is dropped into a coffee cup calorimeter containing 50.0 g?
A piece of copper metal is initially at 100º C. It is dropped into a coffee cup calorimeter containing 50.0 g of water at a temperature of 20.0º C. After stirring, the final temperature of both copper and water is 25.0º C. Assuming no heat losses, and that the specific heat (capacity) of water is 4.184 J(g•K), what is the heat capacity of the copper in J/K? a. 2.79 J/K b. 3.33 J/K c. 13.9 J/K d. 209 J/K e. none of the above
Measuring Heat Capacity of a Calorimeter Cup?
To calibrate your calorimeter cup, you first put 59 mL of cold water in the cup, and measure its temperature to be 18.7 °C. You then pour 51 mL of hot water, temperature = 52.1 °C, into the cup and measure the temperature every thirty seconds over a 10 minute period. You extrapolate this "cooling curve" back to the time of addition and find that the "final temperature" after mixing is 32.3 °C. 1) What is the heat change of the hot water, qHW? (Assume the density of the water is 1.00 g/mL, and remember that the specific heat of water is 4.184 J/g-K or J/g-°C.) 2) What is the heat change of the cold water, qCW? 3) What is the heat change of the calorimeter cup, qcup? 4) What is the heat heat capacity of the calorimeter cup, Ccup? (Enter units either as J/K or J/degC).