Question

What mass of water at 24.0°C must be allowed to come to thermal equilibrium with a 1.90-kg cube of aluminum initially at 150°C to lower the temperature of the aluminum to 69.0°C?

Answer 1

To calculate the mass of water needed to come to thermal equilibrium with the aluminum cube, we can use the principle of heat transfer. The heat lost by the aluminum is equal to the heat gained by the water when it reaches thermal equilibrium. Plugging in the given values, the mass of water required is 1.38 kg.

Step-by-step explanation:

To calculate the mass of water needed to come to thermal equilibrium with the aluminum cube, we can use the principle of heat transfer. The heat lost by the aluminum is equal to the heat gained by the water when they reaches thermal equilibrium. The equation for heat transfer is given by: Q = mcΔT. Where Q is the heat transfer, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. We can rearrange the equation to solve for the mass of water: m(water) = (Q(Al) / (c(water)ΔT(water))). Plugging in the values from the problem, we have: Q(Al) = Q(water). (m(Al) * c(Al) * ΔT(Al)) = (m(water) * c(water) * ΔT(water)). Solving for m(water), we get: m(water) = (m(Al) * c(Al) * ΔT(Al)) / (c(water) * ΔT(water)). Substituting the given values, we have: m(water) = (1.90 kg * 0.897 J/g°C * (69.0°C - 24.0°C)) / (4.18 J/g°C * (150.0°C - 24.0°C)). Calculating this expression gives us: m(water) = 1.38 kg.