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How much heat is required to warm 1.60 kg of sand from 22.0C to 100.0C?
Wiki User
∙ 9y ago
HEAT Q=c*m(t2-t1) c=0.24 (specific heat of sand) t2-t1=100.0oc-20.0oc=80.0oc m=1.40kgQ=0.24*1.40*80.0 Q=26.88 kilocalories or Q=26880 calories
Wiki User
∙ 15y ago
Wiki User
∙ 14y agoq= m Cs (change in temperature)
change in temperature = final temperature - initial temperature ==> 100.0 C- 30.0 C =70C
specific heat capacity of sand (Cs) = .84 J/ g * deg C
mass = 1.7kg ==> convert to grams ==> 1700g
just plug those in to find q which is the amount of heat
q = (1700g) (.84 g/ J *deg C) (70.0 C)
the units should cancel out to give you ....
q =99960 J ==> 100000J (for 3 sig fig if needed)
Wiki User
∙ 9y agoThe heat required, Q , is given by:
Q = ( m ) ( Csh ) ( T2 - T1 )
Q = ( 1.60 kg ) ( 800 J/kg - C ) ( 100.0 C - 22.0 C )
Q = 99800 J <-------
The specific heat of sand, Csh , is a typical value from Heat Transfer Textbook. The
properties of sand can vary considerably, since sand varies a great deal in its composition, moisture content, etc.
Wiki User
∙ 9y agoIt is going to take about 110 kJ to do this heating up. The formula that you will need to use is Q=m*Cp*dT.
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