if it takes 31,500 joules of heat to warm 750g of water, what was the temperature change?
My answer:
Q = (M) (DeltaT) (c)
Delta T=Q
Delta T = 31,500J
--------
750gx 1.0cal/gCo
Delta T = 42oC
1 calorie/gram °C = 4.186 joule/gram °C
you should use 4.186J/g°C not 1.0 cal since the conversion is using grams and energy (J)
If you change that and repost, someone will check it.
You have to multiply it by the specific temperature of water
Well, that's quite a warm water situation you've got there! It seems like the temperature change in this case would be 42 degrees Celsius. I hope the water didn't get too hot and start a tidal wave of laughter!
To find the temperature change, we can use the formula:
Q = (m) (ΔT) (c)
Where:
Q is the amount of heat transferred
m is the mass of the substance (in this case, water)
ΔT is the change in temperature
c is the specific heat capacity of water (which is 1.0 cal/g°C)
We are given that Q = 31,500 J and m = 750 g. The specific heat capacity of water is given in calories per gram per degree Celsius, so we need to convert the units to match.
1 calorie (cal) = 4.184 joules (J)
To convert Q from joules to calories, we divide by the conversion factor:
31,500 J / (4.184 J/cal) = 7,537.87 cal
Now we can substitute the known values into the formula:
7,537.87 cal = (750 g) (ΔT) (1.0 cal/g°C)
ΔT = 7,537.87 cal / (750 g x 1.0 cal/g°C)
ΔT = 42°C
Therefore, the temperature change is 42°C.