What amount of heat is required to raise the temperature of 20 grams of water from 10°C to 30°C? The specific heat of water is 4.18 J/g°C.
Q=MCdeltaT
make sure you change celcius into Kelvin
Q=20(4.18)(20)
Q=1,672
It requires 1,672 Joules
Answer : The amount of heat required is, 1672 J
Solution :
Formula used :
\(Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})\)
where,
Q = heat required = ?
m = mass of water = 20 g
c = specific heat of water = \(4.18J/g^oC\)
\(\Delta T=\text{Change in temperature}\)
\(T_{final}\) = final temperature = \(30^oC\)
\(T_{initial}\) = initial temperature = \(10^oC\)
Now put all the given values in the above formula, we get the amount of heat required.
\(Q=20g\times 4.18J/g^oC\times (30-10)^oC\)
\(Q=1672J\)
Therefore, the amount of heat required is, 1672 J