what is the temperature increase of 6.4 kg of water when it is heated by a 7.7*10 square? (cp=4186j/kg*degrees celsius)
Q = c•m•ΔT
ΔT = Q/ c•m.
Q =7.7*10 square(units???)
To calculate the temperature increase of water when heated, we need to use the specific heat capacity formula:
Q = m * c * ΔT,
where:
Q is the heat energy absorbed or released,
m is the mass of the water,
c is the specific heat capacity of water, and
ΔT is the change in temperature.
In this case, the mass of water is given as 6.4 kg (m), the specific heat capacity is 4186 J/(kg·°C) (cp), and the heat energy absorbed is not specified. We can assume it to be the heat energy provided by the 7.7*10 square.
Now, let's rearrange the formula to solve for ΔT:
Q = m * c * ΔT
ΔT = Q / (m * c)
To find ΔT, we need to know the heat energy provided by the 7.7*10 square. Could you please clarify what you meant by "7.7*10 square"?