Question

Q = cmAT
33. Use the formula to show that 3000 cal are required to
raise the temperature of 300 g of water from 20°C to
30°C. For the specific heat capacity c, use 1 cal/g.°C.

Answer 1

Answer: $Q=3000 cal$

Explanation:

We are given the following formula:

$Q=m. c. \Delta T$   (1)

Where:

$Q=3000 cal$ is the amount of heat

$m=300g$  is the mass  of water

$c=1 cal/g \°C$  is the specific heat of water

$\Delta T$  is the variation in temperature, which in this case is  $\Delta T=30\°C-20\°C=10\°C$  

Rewriting equation (1) with the known values at the right side, we will prove the result is $3000 cal$:

$Q=(300g)(1 cal/g \°C)(10\°C)$   (2)

$Q=3000 cal$   This is the result

Answer 2

The amount of heat required to raise the temperature of water from 20°C to  30°C is 3000 cal. Hence, the statement got verified.

Given data:

The mass of water is, m = 300 g.

The initial temperature of water is, $T_{1}=20^{\circ}\rm C$.

The final temperature of water is, $T_{2}=30^{\circ}\rm C$.

Specific heat capacity of water is, c = 1 cal/g°C.

In this question, we need to verify the end result i,e Q = 3000 cal. Where, Q is the amount of heat required to raise the temperature of water.

And its expression is,

$Q =mc (T_{2} - T_{1})$

Here, c is the specific heat of water.

Solving as,

$Q =300 \times 1 \times (30 - 20)\\\\Q = 3000 \;\rm cal$

Thus, we can conclude that the amount of heat required to raise the temperature of water from 20°C to  30°C is 3000 cal. hence, the statement got verified.

Learn more about the heat capacity here:

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