Question

1. A liquid of mass 250g is heated with an electric heater. Its temperature rises from 30°C to 80°C, the specific heat capacity of the liquid is 2000J/kg°C and the power rating of the heater is 500W. For how long was the heater used?

2. A solid metal of mass 2kg is heated using a heater of power 2kW for 5 minutes. Its final temperature becomes 400°C. Find the initial temperature of the metal. Specific heat capacity of the metal=1J/g°C.

Answer 1

The heater was used for 83.33 minutes to heat the liquid. The initial temperature of the metal was 100°C.

Step-by-step explanation:

To answer the first question, we can use the formula:

Q = mcΔT

Where Q is the heat energy transferred, m is the mass of the liquid, c is the specific heat capacity of the liquid, and ΔT is the change in temperature.

In this case, the mass is 250g, the specific heat capacity is 2000J/kg°C, and the change in temperature is 80°C - 30°C = 50°C. Plugging these values into the formula:

Q = (250g)(2000J/kg°C)(50°C) = 2,500,000J = 2.5MJ

The power rating of the heater is 500W, which means it transfers 500J of energy per second. To find the time it was used, we can divide the heat energy transferred by the power rating:

Time = Q ÷ Power rating = 2.5MJ ÷ 500W = 5000s = 83.33 minutes (rounded to 2 decimal places).

For the second question, we can use the formula:

Q = mcΔT

Where Q is the heat energy transferred, m is the mass of the metal, c is the specific heat capacity of the metal, and ΔT is the change in temperature.

In this case, the power rating of the heater is 2kW, which means it transfers 2000J of energy per second. The time it was used is 5 minutes, or 300 seconds. The change in temperature is 400°C - initial temperature. Plugging these values into the formula and solving for the initial temperature:

Q = (2kg)(c)(400°C - initial temperature) = (2kg)(2000J/g°C)(400°C - initial temperature)

For simplicity, using J/g°C as the unit of specific heat capacity:

2(2000)(400 - initial temperature) = 2(2000)(400 - initial temperature)

800000 - 4000initial temperature = 800000 - 400000

4000initial temperature = 400000

initial temperature = 100°C

Answer 2

1) 50 seconds 2) 100°C

Step-by-step explanation:

(Follows formula of Power=Energy/Time)

1) 500W x X = 2000J/kg°C x .25kg x 50°C

X = 50 seconds.

2) 2000W x 300s = 1000J/kg°C x 2kg x X

X = 300

Initial temperature => 400°C-300°C = 100°C