Question:

A beaker containing 200 mL of water at an initial temperature of 20°C is placed on a hot plate. The hot plate transfers heat to the water at a constant rate of 1000 J/s. The specific heat capacity of water is 4.18 J/g°C. Neglecting any heat loss, calculate the final temperature of the water after 5 minutes of heating.

Answer:

Given: Initial temperature, T₁ = 20°C Volume of water, V = 200 mL = 200 cm³ Rate of heat transfer, Q = 1000 J/s Specific heat capacity of water, c = 4.18 J/g°C

First, we need to determine the mass of water in grams using the given volume: Density of water, d = 1 g/cm³ Mass of water, m = density × volume = 1 g/cm³ × 200 cm³ = 200 g

Next, we can calculate the total heat transferred to the water using the formula: Q = mcΔT, where Q is the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

Since the initial and final temperatures are given, we can write: ΔT = T₂ - T₁

Now, we can rearrange the formula to solve for ΔT: ΔT = Q / mc

Substituting the given values: ΔT = (1000 J/s) / (200 g × 4.18 J/g°C) = 1.196°C

Therefore, the change in temperature of the water is 1.196°C.

Finally, we can find the final temperature, T₂, using the equation: T₂ = T₁ + ΔT

Substituting the values: T₂ = 20°C + 1.196°C = 21.196°C

Hence, the final temperature of the water after 5 minutes of heating is 21.196°C.