How much heat is required to warm 122g of water by 23.0°C? Specific heat is 4.184 J/g° C.
Please show work:(((
heat=mass*specheat*tempchange=122g*4.18J/(gC)*23C
Well, heat is a hot topic, so let's see if we can make it a little more fun! To calculate the amount of heat required, we can use the formula:
Q = m * c * ΔT
Where Q is the heat energy, m is the mass of water, c is the specific heat, and ΔT is the change in temperature. So, let's plug in the values:
m = 122g (mass of water)
c = 4.184 J/g°C (specific heat)
ΔT = 23.0°C (change in temperature)
Now, let's calculate the heat energy:
Q = 122g * 4.184 J/g°C * 23.0°C
Q = 11,998.008 J
Ta-da! The amount of heat required to warm 122g of water by 23.0°C is approximately 11,998.008 J.
To calculate the amount of heat required to warm a given mass of water, you can use the formula:
Q = m * c * ΔT
Where:
Q = heat (in joules)
m = mass of water (in grams)
c = specific heat of water (in J/g°C)
ΔT = change in temperature (in °C)
Now, let's substitute the given values into the formula:
m = 122g
c = 4.184 J/g°C
ΔT = 23.0°C
Q = 122g * 4.184 J/g°C * 23.0°C
Q = 11,470.168 joules
Therefore, the heat required to warm 122g of water by 23.0°C is approximately 11,470.168 joules.
To calculate the amount of heat required to warm the water, you can use the following formula:
Q = m * c * ΔT
Where:
Q = heat energy (in joules)
m = mass of the water (in grams)
c = specific heat capacity of water (in J/g°C)
ΔT = change in temperature (in °C)
Now let's plug in the values:
m = 122g
c = 4.184 J/g°C
ΔT = 23.0°C
Q = 122g * 4.184 J/g°C * 23.0°C
First, multiply the mass (122g) by the specific heat capacity (4.184 J/g°C):
Q = 509.168 J/°C * 23.0°C
Now, multiply this value by the change in temperature (23.0°C):
Q = 11,723.864 J or approximately 11,724 J
Therefore, approximately 11,724 joules of heat are required to warm 122g of water by 23.0°C.