You have 250 g of water in an aluminium calorimeter with a mass of 125 g. How

long will it take to increase the temperature of the water and calorimeter from 20C to 37C if the system is heated by a resistive heating element with 8 V across the resistor and 0.75 A flowing through the resistor?

To answer this question, you need to break it into two parts. Energy = Power*Time (E=P*t). Therefore time = Energy/Power (t=E/P). E and P are what we need to work out:

Power = Voltage*Current (P=V*i)

When heat is added to a solid or a liquid, that energy increases the materials internal energy and thus increases its temperature. The relationship between the heat added and the resulting change in temp is given by:

E=Q(Water) + Q(cup)
Q=m*c*change in temp
Q= heat energy
m= mass
c= specific heat. These aren't given in your question, but are easily found online. c(water)=4184J
and c(aluminum)=900J

So the total energy dissipated as heat = The heat dissipated in the water + heat dissipated in the cup.

E=0.125*900*17 + 0.25*4184*17
=1912.5 + 17782
=19694.5J

Now, we can solve for the final answer:

t= E/P
= 19694.5/6
= 3282.4s
= 54.7 minutes

To calculate the time required to increase the temperature of the water and calorimeter, we need to use the formula for heat transfer:

Q = mcΔT

Where:
Q = heat transfer (in Joules)
m = mass (in kilograms)
c = specific heat capacity (in Joules per kilogram per degree Celsius)
ΔT = change in temperature (in degrees Celsius)

First, we need to calculate the total mass of the water and calorimeter:

Total mass = mass of water + mass of calorimeter

Total mass = 250 g + 125 g = 375 g = 0.375 kg

Next, we need to calculate the total heat transfer:

Q = mcΔT

ΔT = 37°C - 20°C = 17°C

Q = (0.375 kg) × c × 17°C

However, we need the specific heat capacity (c) of the aluminum calorimeter. Let's assume it's 900 J/kg°C.

Q = (0.375 kg) × 900 J/kg°C × 17°C

Now we have the total heat transfer (Q). However, we need to find the time (t) required to reach this temperature increase. To do that, we need to use the formula for electrical power:

P = IV

Where:
P = power (in Watts)
I = current (in Amperes)
V = voltage (in Volts)

The power (P) can also be expressed as:

P = Q / t

Where:
P = power (in Watts)
Q = heat transfer (in Joules)
t = time (in seconds)

We can rearrange the formula to solve for time:

t = Q / P

First, let's calculate the power:

P = IV

P = 8 V × 0.75 A = 6 Watts

Now we can calculate the time required:

t = Q / P

t = (0.375 kg × 900 J/kg°C × 17°C) / 6 W

Now we can calculate the time required to increase the temperature of the water and calorimeter from 20°C to 37°C using the given values.

To determine the time it takes for the temperature of the water and calorimeter to increase from 20°C to 37°C, we need to calculate the amount of heat transferred to the system and then use the specific heat capacity to find the required time.

Step 1: Calculate the heat transferred to the system.
Heat (Q) is given by the formula: Q = mcΔT, where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

The total mass of the system (water + calorimeter) is the sum of the mass of water and the mass of the calorimeter:
Total mass (m) = mass of water + mass of calorimeter = 250 g + 125 g = 375 g = 0.375 kg

The specific heat capacity of water (c) is approximately 4.18 J/g°C.

Using the formula for heat transfer:
Q = mcΔT
Q = (0.375 kg) × (4.18 J/g°C) × (37°C - 20°C)
Q = 0.375 kg × 4.18 J/g°C × 17°C
Q = 26.6635 J

Step 2: Calculate the power input to the system.
Power (P) is given by the formula: P = VI, where V is the voltage and I is the current.

Voltage (V) = 8 V
Current (I) = 0.75 A

Using the formula for power:
P = VI
P = 8 V × 0.75 A
P = 6 W

Step 3: Calculate the time required.
The time (t) is given by the formula: t = Q / P.

Using the values calculated above:
t = 26.6635 J / 6 W
t ≈ 4.44 seconds.

Therefore, it will take approximately 4.44 seconds to increase the temperature of the water and calorimeter from 20°C to 37°C.