how can i get the mass of a liquid when i'm given the heat enrgy that is added, and the specific heat of the liquid to which is added, plus the increase in temperature.

q = mass x specific heat x (Tf - Ti)
q is heat in Joules.
mass in grams
specific heat in J/g*C
Tf = final T
Ti = initial T

specific heat may be in other units, too,

Thanks for your response DrBob222, but i would like to know if to solve mass I only have to solve for "m" like in any algebra equation.

I was given this problem "1200 cal of heat energy is added to a liquid with a specific heat of 0.57cal/gC. If the temperature increases from 20C degree to 33C degree, what is the mass of the liquid?" can you explain how to set it up?

Yes, you solve it as you would any algebra problem. And since q is in calories and specific heat is in calories/g*C, we are ok with units.

1200 cal=q
m = ??
specific heat = 0.57 cal/g*c
Tf = 33
Ti=20
1200=m*0.57*(33-20)
solve for m (in grams)
Check my typing. Check my work.

Thanks I really appreciate your help, I have never gotten to this website before but is great. By the way I got 1192.59 for mass

Check my numbers and check my arithmetic BUT I don't get 1192.59.
I think you must have pushed a wrong button. I have
1200/(0.57*13) = 161.9 g

how many grams of glucose are required to produce 50 mL of0.1 m solution?

sorry am not here to help however i need help i don't know how to do the calorimetry conclusion can anyone help

To solve the problem, you can use the equation:

q = mass x specific heat x (Tf - Ti)

Given:
q = 1200 cal
specific heat = 0.57 cal/g°C
Tf = 33°C
Ti = 20°C

Let's plug in the values and solve for mass (m):

1200 cal = m x 0.57 cal/g°C x (33°C - 20°C)

First, calculate the temperature difference:
33°C - 20°C = 13°C

Now rewrite the equation:

1200 cal = m x 0.57 cal/g°C x 13°C

Next, solve for mass (m):

m = 1200 cal / (0.57 cal/g°C x 13°C)

m ≈ 161.9 g

Therefore, the mass of the liquid is approximately 161.9 grams.

To solve for the mass of the liquid, you can use the formula:

q = mass x specific heat x (Tf - Ti)

Given:

q = 1200 cal
specific heat = 0.57 cal/g*C
Tf = 33C
Ti = 20C

Substituting the values into the formula:

1200 cal = mass x 0.57 cal/g*C x (33C - 20C)

Simplifying the equation:

1200 cal = mass x 0.57 cal/g*C x 13C

Dividing both sides of the equation by (0.57 cal/g*C x 13C):

mass = 1200 cal / (0.57 cal/g*C x 13C)

Evaluating the expression:

mass = 1200 cal / 7.41 cal/g

mass ≈ 161.9 g

Therefore, the mass of the liquid is approximately 161.9 grams.