Question

In a constant-pressure calorimeter, the temperature of 60.0 g of water increases by 4.50 °C. What amount of heat is transferred to the water? (Specific heat capacity of water = 4.2 J/goC) A.7.88 KJ B.1.13 KJ C.1.13 J D.7.88 J

Answer 1

We know that:

- It is a constant-pressure calorimeter

- the temperature of 60.0 g of water increases by 4.50 °C

- Specific heat capacity of water = 4.2 J/goC

And we must find the amount of heat that is transferred to the water.

To find it, we need to use the formula for heat absorbed

$Q=mC_p(T_2-T_1)$

Where,

Q represents the heat absorbed by the water

m represent the mass

Cp represents specific heat, in this case of the water

(T2 - T1) represents the variation of the temperature

Using the given information, we know that:

m = 60.0 g

Cp = 4.2[J/(g°C)]

(T2 - T1) = 4.50 °C

Now, replacing in the formula

$\begin{gathered} Q=60.0g*4.2J/(g°C)*4.50°C \\ Q=1134J \end{gathered}$

Finally, we can convert J to KJ

$Q=1134J=1.13KJ$

ANSWER:

B. 1.13 KJ