Question

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.

Answer 1

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

Answer 2

Q=MCdeltaT
*make sure you change celcius into Kelvin*
Q=20(4.18)(20)
Q=1,672

It requires 1,672 Joules