Question

JC and D

Answer 1

25 degrees celsius will be the final temperature of this sample be after absorbing this energy

Explanation:

data given:

q (heat absorbed) = 990 J

m (mass) = 59 gram

T2 (final temperature) = ?

c  (specific heat capacity of water) = 4.18 J/g °C).

T1 (initial temperature)  = 21

ΔT = T2 -T1

applying the formula for heat absorbed

q = mcΔT

putting the values in the equation

ΔT = $\frac{q}{mc}$

T2 - 21 = $\frac{990}{59 X4.18}$

T2 -21 = 4.01

T2 = 4.01 + 21

     = 25.01 degrees

The final temperature of water sample is 25 degrees when initial temperature was 21 degrees and heat absorbed was 900 J

Answer 2

The final temperature after absorbing the energy of 990 J is 25.014 °C.

Explanation:

The specific heat capacity formula can be used to solve the given problem.

So here the heat absorbed or the Q value is given as 990 J. Then the mass of the sample of water m is given as 59 g. The initial temperature is said to be 21 °C and the final temperature has to be determined.

The specific heat capacity is given as 4.18 J/g°C.

Then, $Q = m*c* del T$

So, the difference in temperature or the change in temperature has to be determined first.

$del T = \frac{Q}{mc} = \frac{990}{59*4.18} = \frac{990}{246.62} \\\\del T =4.014$

So, the change in temperature = ΔT = T₂-T₁ = 4.014 °C

T₂-21 = 4.014

T₂ = 4.014+21 = 25.014 °C.

So, the final temperature after absorbing the energy of 990 J is 25.014 °C.