How do you calculate the amount of heat needed to increase the temperature of 250g of water from 20 C to 56 C? #C_g = 4.8 J/(C-g)#
- Given the data, the formula to use is
#Q=mCpDeltaT#
where:
#m=250g#
#DeltaT=T_2-T_1=56-20=36^oC#
#Cp=(4.8J)/(g*C)# - This case, just plug in given data to find the heat needed to raise the temperature of the water as provided in the problem.
#Q=(250cancel(g))((4.8J)/cancel((g*C)))(36^cancel(oC))#
#Q=43200J#
#Q=43.2kJ#
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To calculate the amount of heat needed, use the formula: Q = mcΔT.
Where:
- Q is the amount of heat needed (in joules)
- m is the mass of the substance (in grams)
- c is the specific heat capacity of the substance (in J/(g°C))
- ΔT is the change in temperature (in °C)
Given:
- m = 250 g
- c = 4.8 J/(g°C)
- ΔT = (56°C - 20°C) = 36°C
Substitute the values into the formula: Q = (250 g) * (4.8 J/(g°C)) * (36°C)
Calculate: Q = 43,200 J
So, the amount of heat needed to increase the temperature of 250g of water from 20°C to 56°C is 43,200 joules.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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