Specific heat means the heat to advanced the temperature of 1 gram that the product by 1 level C.
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So if you want to advanced the temperature of the one gram indigenous 19 to 35 degrees, that exact same gram will require #35-19=16# time as much heat.
And if you desire to advanced the temperature the 193 grams in stead of simply 1 gram, you will need 193 times as much heat.
These two combined give united state the adhering to equation:
#c=#specific heat = because that one gram, one level differencemultiply by:#m=#mass, the lot of grams you desire to warm upmultiply again by:#DeltaT=#the temperature rise you wish
Apr 1, 2015
You know the particular heat the ethanol, which represents the lot of heat required to rise the temperature the 1 gram of ethanol by 1 level Celsius.
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In other words, to get the temperature of 1 gram that ethanol to boost by 1 degree Celsius, you must supply 2.46 J of power to it.
Now, you know the initial and also final temperature of the sample - #19^
"C"# and also #"35"^
"C"#; this means that the temperature of her ethanol sample enhanced by
#DeltaT = T_("final") - T_("initial") = 35 - 19 = 16^
You need to supply enough energy to boost the temperature that the sample through #16^
"C"#, however you have to tak into account how lot ethanol you have, i.e. The fixed given.
The relationship between supplied energy (or heat), mass, and increase in temperature is given by
#q = m * c * DeltaT#, where
#q# - the quantity of power supplied;#m# - the massive of the problem - in your case, the fixed of ethanol;#c# - the specific heat of ethanol;#DeltaT# - the adjust in temperature;
To recognize how lot energy is required to heat 193 g of ethanol through #16^
"C"#, plug her values into the above equation
#q = "193"cancel("g") * 2.46"J"/(cancel("g") * ^
cancel("C")) * 16^
cancel("C") = "7596.48 J"#
Rounded to 2 sig figs, the number of sig figs offered for 19 and 35 levels Celsius, the answer will certainly be
#q = color(green)("7600 J")#
Apr 1, 2015
The heat compelled is same to the mass of ethanol (193 g) multiply by the details heat capacity (2.46 J/g-C) times the preferred temperature difference (35 C - 19 C = 16 C):
#193 g time 2.46 J/(g-C) time 16C=7596 J#
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