# IBDP Physics 3.1 – Thermal concepts: IB Style Question Bank : HL Paper 1

### Question

A container with 0.60kg of a liquid substance is placed on a heater at time t=0. The specific latent heat of vaporization of the substance is 200kJkg–1. The graph shows the variation of the temperature T of the substance with time t. What is the power of the heater?

A. 1200 W

B. 3000 W

C. 4800 W

D. 13 300 W

### Markscheme

B

We know $$Power =P=\frac{\Delta W}{\Delta t}$$

Now $$\Delta W =mL= 0.6 \times 200 \times 10^3$$
and $$\Delta t = (50-10) =40 s$$
Hence $$\Delta P=\frac{0.6 \times 200 \times 10^3}{40} =3000W$$

### Question

An insulated tube is filled with a large number n of lead spheres, each of mass m. The tube is inverted s times so that the spheres completely fall through an average distance L each time. The temperature of the spheres is measured before and after the inversions and the resultant change in temperature is ΔT.

What is the specific heat capacity of lead? A $$\frac{sgL}{nm\Delta T}$$

B $$\frac{sgL}{\Delta T}$$

C $$\frac{sgL}{n\Delta T}$$

D $$\frac{gL}{m\Delta T}$$

Ans: B

Here Mechanical energy is converted into Heat Energy.

Each Sphere is moved through a distance of L . hence Gravitation field energy for each sphere each time of inversion = $$mgL$$

Now there are $$n$$ sphere which are inverted $$s$$ number of times . Hence mechanical (gravitation energy) = $$nsmgL$$

Now $$n$$ sphere temperature raise by $$\Delta T$$ times. Let $$c$$ be specific heat capacity of lead

Hence total heat energy required to raise temperature by $$\Delta T$$ = $$nmc\Delta T$$

Hence

$$nmc\Delta T = nsmgL$$

or

$$c=\frac{nsmgL}{nm\Delta T}=\frac{sgL}{\Delta T}$$

### Question

A mass m of water is at a temperature of 290 K. The specific heat capacity of water is c . Ice, at its melting point, is added to the water to reduce the water temperature to the freezing point. The specific latent heat of fusion for ice is L . What is the minimum mass of ice that is required?

A $$\frac{17mc}{L}$$

B $$\frac{290mc}{L}$$

C $$\frac{17mL}{c}$$

D $$\frac{290mL}{c}$$

$$mc(290-273)={m}’L$$
$$17mc={m}’L$$
$${m}’=\frac{17mc}{L}$$