Physics: The specific heat of water is 4.2 j/c. if it takes 31,500 joules to heat to warm 750 g of water, what was the temperature change?

The specific heat of water is 4.2 j/c. if it takes

Ответ:

kiyomi58

14.03.2022

The amount of heat needed to increase the temperature of a substance by $\Delta T$ is given by
$Q=mC_s \Delta T$
where
m is the mass of the substance
C_s

the specific heat capacity
$\Delta T$ the increase in temperature

In our problem, the mass of the water is m=750 g, the specific heat is $C_s = 4.2 J/g ^{\circ}C$ and the amount of heat supplied is Q=31500 J

, so if we re-arrange the previous formula we find the increase in temperature of the water:
$\Delta T= \frac{Q}{m C_s}= \frac{31500 J}{(750 g)(4.20 J/g^{\circ} C)}=10^{\circ}C$

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Ответ:

adeanbates

15.05.2020

Physics

Find the rate of change:

(65,165) , (75,195)

(195 - 165)/(75-65)

30/10

3 pounds average rate of change

58.3*3 =174.9

Hope this helps!

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The specific heat of water is 4.2 j/c. if it takes 31,500 joules to heat to warm 750 g of water, what was the temperature change?

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