If the temperature of 100 g of water increases from 10 to 45 C, how much heat is absorbed? (Specific heat of water = 4.184 J/gC)
The heat absorbed can be calculated using the formula:
Q = m * c * ΔT
Where:
Q = heat absorbed
m = mass of the water (100 g)
c = specific heat of water (4.184 J/gC)
ΔT = change in temperature (45 - 10 = 35 C)
Substituting the values into the formula:
Q = (100 g) * (4.184 J/gC) * (35 C)
Q = 14644 J
Therefore, 14644 J of heat is absorbed.
To calculate the amount of heat absorbed by the water, we can use the formula:
Q = m * c * ΔT
Where:
Q is the amount of heat absorbed (in joules)
m is the mass of the water (in grams)
c is the specific heat of water (in J/g°C)
ΔT is the change in temperature (in °C)
Given:
m = 100 g
c = 4.184 J/g°C
ΔT = 45°C - 10°C = 35°C
Now we substitute the values into the formula:
Q = 100 g * 4.184 J/g°C * 35°C
Q = 14644 J
Therefore, the amount of heat absorbed is 14644 joules.
To calculate the amount of heat absorbed by the water, you can use the formula:
Q = m * c * ΔT
Where:
Q is the heat absorbed (in Joules)
m is the mass of the water (in grams)
c is the specific heat capacity of water (in J/g•°C)
ΔT is the change in temperature (in °C)
In this case, the mass of water (m) is 100 g, the specific heat capacity of water (c) is 4.184 J/g•°C, and the change in temperature (ΔT) is 45°C - 10°C = 35°C.
Plugging these values into the formula, we get:
Q = 100 g * 4.184 J/g•°C * 35°C
To solve this equation, we can multiply the values:
Q = 14644 J
Therefore, the amount of heat absorbed by the water is 14644 Joules.