# Homework Help: Specific and Latent Heat - Pre-laboratory work

1. May 16, 2008

### CGUE

I just need any of you to check if I've answered the problems correctly, thanks.

The problem statement, all variables and given/known data

Question 1.

A student uses some equipment that can add energy at a constant rate to some water contained in a copper calorimeter can. To keep the temperature of the water uniform, the student stirs the water with a copper stirrer. Together, the calorimeter can and the stirrer have a mass of mc and a specific heat of cc and the water has a mass of mw and specific heat of cw.

The initial temperature of the water is T0, and the student begins adding energy to the system at a constant rate of P watts.

Assuming there is no energy loss from the system, derive an expression for the temperature, T, of the system as a function of time, t.

(Hint: the rate at which the energy is added, P = $$\frac{dQ}{dt}$$ where dQ = (mccc+mwcw)dT.
Substitute,rearrange and integrate).

Question 2.

Consider a copper calorimeter can with a copper stirrer has a total mass mc and specific heat cc.
• It contains a msss mw of water of specific heat cw at a temperature of TioC.
• A mass mi of ice at 0oC is added to the calorimeter.
• The heat of fusion of ice is Li.
• All the ice melts and the temperature of the can, stirrer, and contents falls to TfoC (which is above 0oC)
(a) Presuming that there is no heat energy gained or lost by the system, write down an equation representing conservation of energy for the system. Indicate clearly what each terms in this equation represents.
(b) Hence, find an expression for the latent heat of fusion of ice, Li, in terms of the other variables in your equation. (Show all steps)

The attempt at a solution

Question 1

P = $$\frac{dQ}{dt}$$ = $$\frac{(m_cc_c+m_wc_w)dT}{dt}$$

dT = $$\frac{P dt}{m_cc_c+m_wc_w}$$

T = $$\frac{P}{m_cc_c+m_wc_w}$$$$\int_{t_i}^{tf} dt$$

T = $$\frac{P}{m_cc_c+m_wc_w}$$$$\left[ t \right]_{t_i}^{t_f}$$

T = $$\frac{P(t_f-t_i)}{m_cc_c+m_wc_w}$$

Question 2 a

Qi = -Qw+c
Qi - Qw+c = 0
miLi + mw+ccw+c$$\Delta$$T = 0
miLi + (mccc+mwcw)(Tf - Ti) = 0

Question 2 b

miLi + Tf(mccc+mwcw) - Ti(mccc+mwcw) = 0

miLi = Ti(mccc+mwcw) - Tf(mccc+mwcw)

.: Li = $$\frac{(T_i-T_f)(m_cc_c+m_wc_w)}{m_i}$$

Last edited: May 16, 2008