Consider 30.0 mL of water that has absorbed 1540 J of heat from a chemical reaction. What is the change in temperature that should be observed in degrees Celsius?
Delta t =q/mc *1.00g
The 1.00g is to convert the ml to g in order to cancel out what u don’t need
12.3 degrees Celsius
do we have the same lab?
u know how to do the third one?
how'd you get 12.3 degrees C
that answer you got is in K not degrees C
Its in Celcius because the specific heat of water is 4.184J/gºC so the equation will leave you with ºC
To find the change in temperature, we can use the specific heat capacity formula:
q = m * c * ΔT
Where:
- q is the heat absorbed (in Joules),
- m is the mass of the substance (in grams),
- c is the specific heat capacity of the substance (in J/g·°C), and
- ΔT is the change in temperature (in °C).
In this case, we are given:
- q = 1540 J
- m = mass of water = 30.0 g
- c = specific heat capacity of water = 4.18 J/g·°C
Plugging in these values into the formula, we get:
1540 J = 30.0 g * 4.18 J/g·°C * ΔT
Now, we can solve for ΔT:
ΔT = 1540 J / (30.0 g * 4.18 J/g·°C)
Calculating this, we get:
ΔT ≈ 15.59 °C
Therefore, the change in temperature observed should be approximately 15.59 degrees Celsius.