an 800kg camel starts out this extremely hot day with a body temperature of 36C.
ID: 1326597 • Letter: A
Question
an 800kg camel starts out this extremely hot day with a body temperature of 36C. By the days end, this camel temperature has risen to 44C. A. How much get was stored by the camel allowing it's temperature to rise? B. If the camel had defended a Tb of 36 degrees C how much water would he have had to use for evaporative cooling? Specific heat of camel. = 0.9 cal/g*deg C Heat of vaporization of water = .584 kcal/g an 800kg camel starts out this extremely hot day with a body temperature of 36C. By the days end, this camel temperature has risen to 44C. A. How much get was stored by the camel allowing it's temperature to rise? B. If the camel had defended a Tb of 36 degrees C how much water would he have had to use for evaporative cooling? Specific heat of camel. = 0.9 cal/g*deg C Heat of vaporization of water = .584 kcal/g an 800kg camel starts out this extremely hot day with a body temperature of 36C. By the days end, this camel temperature has risen to 44C. A. How much get was stored by the camel allowing it's temperature to rise? B. If the camel had defended a Tb of 36 degrees C how much water would he have had to use for evaporative cooling? Specific heat of camel. = 0.9 cal/g*deg C Heat of vaporization of water = .584 kcal/gExplanation / Answer
a)
Heat Gained by camel = Mass of camel*specif heat of camel*(change in temperature of camel)
= 800 * 0.9*(1000) * (44-36) = 5760000 calories
b) Heat gained by camel = heat lost for evaporative cooling
so, Heat gained by camel = 5760000 calories ----------1)
heat lost for evaporative cooling = Mass of water * Heat of vaporization of water
= Mass of water * .584*(1000) * (1000) calories -------------2)
equating 1 and 2 we get
Mass of water * .584*(1000) * (1000) calories = 5760000 calories
Mass of water required = 9.36 Kg
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