A 700-g insulated metal dish contains 1000 g of water at 20.00°C . A 500-g iron bar at 120°C is dropped into the water, and the final temperature is 24.77°C .

Question

A 700-g insulated metal dish contains 1000 g of water at 20.00°C . A 500-g iron bar at 120°C is dropped into the water, and the final temperature is 24.77°C . What is the specific heat of the metal dish? cwater = 4.19 J / g C°, ciron 0=.447 J / g C . °

Answer 1

Q = c*m*ΔT

You have a heat transfer from a hot object (iron bar) to the cold object (metal dish and water).

Since the system is insulated, no heat nor mass is transferred to the environment, meaning the hot iron bar will give off heat energy only to the dish with water until equilibrium is reached (all at equal temperature).

The amount of heat energy that left the iron bar (Q_H) equals the amount of heat absorbed by the dish and water (Q_C).

Q_H = Q_C

c_iron*m_iron*ΔT_iron = (c_water*m_water + c_dish*m_dish)*ΔT_C,

(ΔT_C = ΔT_water = ΔT_dish, since both were at equilibrium before adding the iron bar and at the end).

Solve the above equation for c_dish.

c_dish = (c_iron*m_iron*ΔT_iron / ΔT_C - c_water*m_water) / m_dish

It's a zinc dish.