The specific heat of water is 4186 J/(kg·°C)
(40 x 10^3 J) / (4186 J/(kg·°C)) / (4 kg) = 2.39 °C change
30°C + 2.39 °C = 32 °C
the equation is
q = mSdeltaT
q is heat added, m is mass in g, S the specific heat, and deltaT the change in temp.
Calculate deltaT and add it to 30 C
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Answers & Comments
(40 x 10^3 J) / (4186 J/(kg·°C)) / (4 kg) = 2.39 °C change
30°C + 2.39 °C = 32 °C
the equation is
q = mSdeltaT
q is heat added, m is mass in g, S the specific heat, and deltaT the change in temp.
Calculate deltaT and add it to 30 C