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
Good
To find the temperature change (ΔT), we can use the equation Q = (M) (ΔT) (c), where Q is the heat energy transferred, M is the mass of the substance, ΔT is the temperature change, and c is the specific heat capacity of the substance.
In this case, we are given that it takes 31,500 joules (J) of heat to warm 750 grams (g) of water. We know the mass (M) and the heat energy (Q), so we can rearrange the equation to solve for ΔT.
ΔT = Q / (M * c)
First, we need to convert the mass from grams to kilograms, as the specific heat capacity is usually given in J/(g·K). So, 750 grams = 750/1000 = 0.75 kilograms (kg).
The specific heat capacity of water is approximately 1.0 cal/g·°C. To convert from calories to joules, we can use the conversion factor: 1 cal = 4.184 J.
Now, we can calculate the temperature change:
ΔT = 31,500 J / (0.75 kg * 1.0 cal/g·°C * 4.184 J/cal)
Simplifying the equation:
ΔT = 31,500 J / (0.75 kg * 4.184 J/g·°C)
ΔT = 42 °C
Therefore, the temperature change of the water is 42 °C.