Question

What amount of heat is absorbed by 257 g of water heated from 18° C to 63° C? The specific heat of water is 4.18 J/g°C.

Answer 1

Answer:

Q = 48341.7 Joules.

Explanation:

Given the following data;

Mass, m = 257g

Initial temperature, T1 = 18°C

Final temperature, T2 = 63°C

Specific heat capacity of water, c = 4.18 J/g°C.

*To find the quantity of heat absorbed*

Heat capacity is given by the formula;

$Q = mcdt$

Where;

• Q represents the heat capacity or quantity of heat.

• m represents the mass of an object.

• c represents the specific heat capacity of water.

• dt represents the change in temperature.

dt = T2 - T1

dt = 63 - 18

dt = 45°C

Substituting the values into the equation, we have;

$Q = 257*4.18*45$

Q = 48341.7 Joules.

Therefore, the amount of heat absorbed is 48341.7 Joules.