Physics problem about a sphere in space

Question

In open space, a sphere of radius R is filled with a liquid whose initial

temperature T0 (higher than its enviroment). Find the dependence of temperature of the liquid with time. The mass of the liquid is m, and the specific heat capacity is c. The sphere is painted black.

Answer 1

The temperature of the liquid will decrease with time due to heat loss to the environment. The rate of heat loss is proportional to the temperature difference between the liquid and the environment, according to Newton's law of cooling.

Let T(t) be the temperature of the liquid at time t. Then, the rate of heat loss is given by:

dQ/dt = -kA(T(t) - T_env)

where dQ/dt is the rate of heat loss, k is the thermal conductivity of the liquid, A is the surface area of the sphere, and T_env is the temperature of the environment.

The rate of heat loss is also equal to the rate of change of thermal energy of the liquid:

dQ/dt = mc dT/dt

where m is the mass of the liquid and c is its specific heat capacity.

Equating the two expressions for dQ/dt, we get:

mc dT/dt = -kA(T(t) - T_env)

Simplifying and separating variables, we get:

dT/(T(t) - T_env) = -kA/(mc) dt

Integrating both sides, we get:

ln(T(t) - T_env) = -kA/(mc) t + C

where C is the constant of integration.

Solving for T(t), we get:

T(t) = T_env + exp(-kA/(mc) t + C)

We can determine the value of the constant C from the initial condition T(0) = T0:

T(0) = T_env + exp(C) = T0

Therefore, C = ln(T0 - T_env).

Substituting C into the expression for T(t), we get:

T(t) = T_env + (T0 - T_env) exp(-kA/(mc) t)

This is the dependence of the temperature of the liquid with time. It shows that the temperature exponentially approaches the temperature of the environment as time passes.