Imagine that you have a heater (similar to the one used in the lab) containing m
ID: 1474080 • Letter: I
Question
Imagine that you have a heater (similar to the one used in the lab) containing m = 529g of unknown liquid. You heat the liquid up by using a heat source of constant power P = 459W and measure the temperature of liquid measured in degree C as function of time. Assume there is no energy loss to the environment during the heating process and the heating occurred below the boiling point of the liquid. The graph of temperature vs. time in this experiment will resemble which of the following graphs the most? You measured the slope of the resulting curve in the temperature range close to the room temperature to be slope= 0.181 degree C/s . What is the specific heat of the unknown liquid? Assume you heated the unknown liquid from the previous question from room temperature to 90 degree C (assume that the unknown liquid is still not boiling). Compared with the same amount of water using the same heater (say you were cooking pasta in that unknown liquid)Explanation / Answer
Here ,
P * time = m * Specific heat * change in temperature
P * time = m * Specific heat * change in temperature
hence , the temperature is directly porporational to the time
hence , the correct graph is
bottom left graph
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mass , m = 529 gm = 0.529 Kg
P * time = m * Specific heat * change in temperature
change in temperature = (P/(mass * specicfic heat )) * time
hence , slope = (P/(mass * specicfic heat ))
0.181 degree C/s = 459/(0.529 * specicfic heat )
specicfic heat = 4793.8 J/(kg * C)
specicfic heat = 4.79 kJ/(Kg * degree C )
the specific heat of unknown liquid is 4.79 kJ/(Kg * degree C )
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Here, specific heat of water = 4.186 kJ/(Kg.degree C)
as the specific heat of water is less than the specific heat of unknown liquid.
for changing the same temperature
hence , for the same heater , water will take less heat
the correct option is
5) it will take longer time because the specific heat of unknown liquid is larger than water