How much of heat is needed to warm 500 g of water from 25 °C near to its boiling point of 99°C? The specific heat of capacity is 4.18J/g-k.
2.10 X10 squared J
1.55 X 10 to the power of 5 J
67.9 J
52.3 J
The correct answer is 1.55 x 10^5 J.
To calculate the amount of heat needed, we need to use the formula:
Q = m * c * ΔT
where Q is the amount of heat, m is the mass of the water, c is the specific heat capacity, and ΔT is the change in temperature.
In this case, we have:
m = 500 g
c = 4.18 J/g-K
ΔT = (99°C - 25°C) = 74°C
Plugging in these values, we get:
Q = 500 g * 4.18 J/g-K * 74°C
Q = 155,340 J
Q ≈ 1.55 x 10^5 J
Therefore, we need 1.55 x 10^5 J of heat to warm 500 g of water from 25°C to its boiling point of 99°C.
To calculate the amount of heat needed to warm the water, we first need to determine the temperature change. The initial temperature is 25 °C and the final temperature is 99 °C. So the temperature change is:
ΔT = final temperature - initial temperature
ΔT = 99 °C - 25 °C
ΔT = 74 °C
Next, we can calculate the amount of heat using the formula:
Q = m * c * ΔT
Where:
Q is the amount of heat
m is the mass of the water (500 g)
c is the specific heat capacity (4.18 J/g-°C)
ΔT is the temperature change (74 °C)
Substituting the given values into the formula:
Q = 500 g * 4.18 J/g-°C * 74 °C
Q = 1.55 x 10^5 J (rounded to two significant figures)
Therefore, the amount of heat needed is approximately 1.55 x 10^5 J.