Chemistry Help?

Question

The specific heat of water is 4.184 J/g*C degree. How much heat is required ot raise the temperature of 5.0 g of water by 3.0 C

A 21.0g sample of water is cooled from 34.0 degree celcius to 28 degree celcius. How many jouoles of heeat were removed from teh water.

I have no idea how to do this so could someone help me out

Answer 1

Let the units be your guide...

5 g * 4.184 J/(g*C) * 3 C = 62.76 J

21 g * 4.184 J/(g*C) * 6 C = 527.2 J

Answer 2

The needed formula for both these questions is:
Q = mc(delta T)
Where Q is the heat energy which is added / taken away from the water, m is the mass of the water, c is the specific heat of the water, and (delta T) is the change in temperature the water experiences.

You are given the mass of the water in both questions and you know the specific heat. You are also given either the change in temperature of the initial and final temperatures of the water (from which you can calculate the change). You have all the information you need to solve for the heat energy (Q).

In the first problem,
m = 5.0 g
(delta T) = 3 degrees C,
Q = (5.0 g) * (4.184 J / g degree C) * (3.0 degrees C)
Q = 62.75 J
Q = 63 Joules are absorbed by the water

In the second problem,
m = 21.0 g
(delta T) = 28 degrees C - 34 degrees C = -6 degrees C,
Q = (21.0 g) * (4.184 J / g degree C) * (-6.0 degree C)
Q = -527.184 J
Q = 530 Joules released by the water.
(The negative sign on the heat energy indicates that the energy is flowing out of the water [exothermic process], you can often ignore the sign if you remember to keep the direction of heat flow straight by some other means).

Answer 3

deltaQ=mcdeltaT

change in heat=mass x specific heat x change in temp.

deltaQ=5.0gx4.184J/g*Cx3.0 C
deltaQ=62.76 J
sig fig=63


deltaQ=21.0gx4.184J/g*Cx6 C
deltaQ=527.184 J
sig fig=500 J

Sig figs, in addition and subtraction, go on the place value instead of the number of significant digits.. so in 34.0-28 the answer would be 6 not 6.0, making the final answer 500 J not 530 J.